Contents / Calculus / Techniques of Integration

Chapter 5

Techniques of Integration

Integration by parts, trigonometric integrals and substitutions, partial fractions, improper integrals, and choosing a strategy.

Introduction

Differentiation is mechanical. The product rule, the quotient rule and the chain rule let you differentiate any expression built from the standard functions, and the answer is again built from the standard functions. Integration has no such guarantee. There is no product rule for integrals, no chain rule, and the antiderivative of a harmless-looking function such as $e^{-x^{2}}$ cannot be written down in terms of elementary functions at all.

What there is instead is a collection of techniques, each of which converts one recognisable shape of integrand into something the basic table can handle. Integration by parts deals with products. Trigonometric integrals and trigonometric substitution deal with powers of trigonometric functions and with square roots of quadratics. Partial fractions deal with rational functions. Rationalising substitutions deal with roots of linear expressions. The chapter closes with improper integrals, where the interval is infinite or the integrand blows up, and with a strategy for deciding, when you meet an integral in the wild, which technique to reach for first.

Everything rests on the table from the Integrals chapter: powers, $e^x$ , $\ln x$ , the six trigonometric derivatives read backwards, $\arcsin$ and $\arctan$ , together with the substitution rule. Each technique below is a way of transforming an integral until it lands in that table.

5.1Integration by parts

Theorem 5.1 (Integration by parts). If $f$ and $g$ are differentiable and $f'g$ and $fg'$ are continuous, then

\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int g(x)\,f'(x)\,dx.

In the shorthand $u = f(x)$ , $dv = g'(x)\,dx$ , $du = f'(x)\,dx$ , $v = g(x)$ this reads

\int u\,dv = uv - \int v\,du.

Proof. The product rule says $\frac{d}{dx}\big[f(x)g(x)\big] = f'(x)g(x) + f(x)g'(x)$ . Integrating both sides,

f(x)g(x) = \int f'(x)g(x)\,dx + \int f(x)g'(x)\,dx,

and moving the first integral to the other side gives the formula. The constant of integration is absorbed into the remaining indefinite integral.∎

Corollary 5.2 (Integration by parts for definite integrals). Under the same hypotheses on $[a,b]$ ,

\int_a^b f(x)\,g'(x)\,dx = \Big[f(x)\,g(x)\Big]_a^b - \int_a^b g(x)\,f'(x)\,dx.

The formula does not compute the integral; it trades it for a different one, $\int v\,du$ , plus the boundary term $uv$ . The trade is worthwhile exactly when $\int v\,du$ is easier than $\int u\,dv$ , and that depends entirely on how you split the integrand. Two things change between the two integrals: $u$ is replaced by its derivative $du$ , and $dv$ is replaced by its antiderivative $v$ . So the ideal $u$ is something that gets simpler when differentiated, and the ideal $dv$ is something you can integrate without getting more complicated.

Polynomials get simpler under differentiation: $x^{3}$ becomes $3x^{2}$ , and after three rounds becomes a constant. Logarithms and inverse trigonometric functions get dramatically simpler: $\ln x$ becomes $1/x$ , $\arctan x$ becomes $1/(1+x^{2})$ , both algebraic. Exponentials and $\sin$ , $\cos$ do not get simpler under differentiation, but they do not get worse under integration either, which makes them ideal for $dv$ . The mnemonic LIATE orders the families by how much they want to be $u$ .

Method 5.3 (Integration by parts with LIATE).

Write the integrand as a product $u \cdot dv$ , where $dv$ includes $dx$ . To choose $u$ , take the factor that comes first in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Everything else is $dv$ .
Compute $du$ by differentiating $u$ and $v$ by integrating $dv$ . Do not add a constant to $v$ ; any antiderivative works, and the simplest is best.
Write $uv - \int v\,du$ and evaluate the new integral. If it needs parts again, keep the same roles: whatever was differentiated stays on the differentiating side.
Check by differentiating your answer; the product rule should return the original integrand.

LIATE is a heuristic, not a theorem, and it has limits. It assumes $dv$ can actually be integrated: in $\int x^{3} e^{x^{2}}\,dx$ the exponential $e^{x^{2}}$ has no elementary antiderivative, so $dv = e^{x^{2}}\,dx$ is useless. The right split is $u = x^{2}$ and $dv = x e^{x^{2}}\,dx$ , because $\int x e^{x^{2}}\,dx = \tfrac12 e^{x^{2}}$ by substitution. When the integrand contains a single function, as in $\int \ln x\,dx$ or $\int \arcsin x\,dx$ , the missing factor is $dv = dx$ , and parts works because $du$ is algebraic.

Intuition. Take $\int_0^1 x e^x\,dx$ and think of the derivative as a burden that one of the two factors must carry. On its own, $x e^x$ is not the derivative of anything obvious. Parts moves the burden: differentiate the $x$ (it becomes $1$ , which is as light as a factor gets) and integrate the $e^x$ (it stays $e^x$ , no heavier). The price is the boundary term $\big[x e^x\big]_0^1 = e - 0 = e$ , and what remains is the light integral $\int_0^1 e^x\,dx = e - 1$ . The total is $e - (e - 1) = 1$ . The burden was moved onto the factor that could carry it.

Example 5.4 (A polynomial times an exponential). Compute $\int x e^{2x}\,dx$ .

Solution.

The factors are algebraic ( $x$ ) and exponential ( $e^{2x}$ ). By LIATE, $u = x$ and $dv = e^{2x}\,dx$ .
Then $du = dx$ and $v = \int e^{2x}\,dx = \tfrac12 e^{2x}$ .
The formula gives

\int x e^{2x}\,dx = \frac{x e^{2x}}{2} - \int \frac{e^{2x}}{2}\,dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C.

Check: differentiating $\frac{x e^{2x}}{2}$ by the product rule gives $\frac{e^{2x}}{2} + x e^{2x}$ , and differentiating $-\frac{e^{2x}}{4}$ gives $-\frac{e^{2x}}{2}$ . The sum is $x e^{2x}$ , as required.

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Example 5.5 (A single function: the $dv = dx$ trick). Compute $\int \arctan x\,dx$ .

Solution.

There is only one factor, an inverse trigonometric function, which wants to be $u$ . Take $u = \arctan x$ and $dv = dx$ .
Then $du = \frac{dx}{1+x^{2}}$ and $v = x$ .
By parts,

\int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx.

The remaining integral is a substitution: with $w = 1 + x^{2}$ , $dw = 2x\,dx$ , it equals $\tfrac12 \int \frac{dw}{w} = \tfrac12 \ln(1+x^{2})$ . No absolute value is needed because $1 + x^{2} > 0$ .
So $\int \arctan x\,dx = x\arctan x - \tfrac12\ln(1+x^{2}) + C$ .
Check: the derivative of $x\arctan x$ is $\arctan x + \frac{x}{1+x^{2}}$ , and the derivative of $-\tfrac12\ln(1+x^{2})$ is $-\frac{x}{1+x^{2}}$ . They combine to $\arctan x$ .

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The same move handles $\int \ln x\,dx = x\ln x - x + C$ and $\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^{2}} + C$ : in each case $du$ is algebraic and the leftover integral is elementary.

Example 5.6 (Parts applied twice). Compute $\int x^{2} \sin x\,dx$ .

Solution.

Take $u = x^{2}$ (algebraic beats trigonometric) and $dv = \sin x\,dx$ , so $du = 2x\,dx$ and $v = -\cos x$ .
First application:

\int x^2 \sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx.

The new integral still has a polynomial factor, but of lower degree. Apply parts again with $u = x$ , $dv = \cos x\,dx$ , so $du = dx$ , $v = \sin x$ :

\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x.

Substituting back,

\int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + C.

Check: differentiate. From $-x^{2}\cos x$ comes $-2x\cos x + x^{2}\sin x$ ; from $2x\sin x$ comes $2\sin x + 2x\cos x$ ; from $2\cos x$ comes $-2\sin x$ . Everything cancels except $x^{2}\sin x$ .

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When the polynomial factor has degree $n$ , parts must be applied $n$ times, always differentiating the polynomial. The tabular method organises the bookkeeping: list the successive derivatives of the polynomial in one column ( $x^{2}$ , $2x$ , $2$ , $0$ ) and the successive antiderivatives of the other factor in a second column ( $\sin x$ , $-\cos x$ , $-\sin x$ , $\cos x$ ), then multiply diagonally with alternating signs $+,-,+,\ldots$ . For the example above this produces $x^{2}(-\cos x) - 2x(-\sin x) + 2(\cos x)$ , the same answer in one line.

Example 5.7 (The boomerang: an exponential times a trigonometric function). Compute $\int e^x\cos x\,dx$ .

Solution.

Neither factor simplifies under differentiation, so LIATE puts $u = \cos x$ (trigonometric) and $dv = e^x\,dx$ . Then $du = -\sin x\,dx$ and $v = e^x$ . Call the integral $I$ :

I = e^x\cos x + \int e^x\sin x\,dx.

Apply parts to the new integral with the same roles, $u = \sin x$ , $dv = e^x\,dx$ , $du = \cos x\,dx$ , $v = e^x$ :

\int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx = e^x\sin x - I.

The original integral has come back. Substitute it into step 1:

I = e^x\cos x + e^x\sin x - I.

This is an equation for $I$ . Solve it: $2I = e^x(\cos x + \sin x)$ , so

\int e^x\cos x\,dx = \frac{e^x(\sin x + \cos x)}{2} + C.

Check: the derivative of $\tfrac12 e^x(\sin x + \cos x)$ is $\tfrac12 e^x(\sin x + \cos x) + \tfrac12 e^x(\cos x - \sin x) = e^x\cos x$ .

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The trick works because after two rounds the trigonometric factor returns to itself up to sign, and the sign is what makes the equation solvable. Had it returned with a plus sign, the two sides would cancel to $0 = 0$ . It is essential to keep the same roles in both rounds: if in step 2 you take $u = e^x$ instead, you simply undo step 1.

Example 5.8 (A definite integral by parts). Evaluate $\int_0^{1/2} \arcsin x\,dx$ .

Solution.

Take $u = \arcsin x$ , $dv = dx$ , so $du = \frac{dx}{\sqrt{1-x^{2}}}$ and $v = x$ .
The definite form of the theorem gives

\int_0^{1/2}\arcsin x\,dx = \Big[x\arcsin x\Big]_0^{1/2} - \int_0^{1/2}\frac{x}{\sqrt{1-x^2}}\,dx.

The boundary term is $\tfrac12\arcsin\tfrac12 - 0 = \tfrac12\cdot\tfrac{\pi}{6} = \tfrac{\pi}{12}$ .
For the remaining integral substitute $w = 1 - x^{2}$ , $dw = -2x\,dx$ ; the limits become $w = 1$ and $w = \tfrac34$ :

\int_0^{1/2}\frac{x}{\sqrt{1-x^2}}\,dx = -\frac12\int_1^{3/4} w^{-1/2}\,dw = -\frac12\Big[2\sqrt{w}\Big]_1^{3/4} = -\Big(\frac{\sqrt3}{2} - 1\Big) = 1 - \frac{\sqrt3}{2}.

Therefore

\int_0^{1/2}\arcsin x\,dx = \frac{\pi}{12} - 1 + \frac{\sqrt3}{2} \approx 0.128.

Sanity check: on $[0, \tfrac12]$ we have $\arcsin x \ge x$ , so the integral should slightly exceed $\int_0^{1/2} x\,dx = \tfrac18 = 0.125$ . It does.

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Pitfall. The formula is $uv - \int v\,du$ , with a minus sign. Dropping it is the most common error, and it is invisible until you check by differentiating. A second trap is choosing $dv$ to be something you cannot integrate: $dv = \ln x\,dx$ or $dv = \arctan x\,dx$ are always wrong choices, because $v$ would be harder than what you started with.

Remark (A false paradox). Apply parts to $\int \frac{1}{x}\,dx$ with $u = \frac1x$ , $dv = dx$ , so $du = -\frac{dx}{x^{2}}$ , $v = x$ :

\int \frac{1}{x}\,dx = \frac1x\cdot x - \int x\cdot\Big(-\frac{1}{x^2}\Big)dx = 1 + \int\frac{1}{x}\,dx,

which seems to say $0 = 1$ . The resolution is that an indefinite integral is a family of functions, not a single one, and the two occurrences of $\int \frac1x\,dx$ carry different constants. The equation says $\ln|x| + C_1 = 1 + \ln|x| + C_2$ , which is true. It is the same reason the boomerang trick must be written with a single named $I$ and solved as an equation, with $+C$ added only at the very end.

Exercises for §5.1 →

5.2Reduction formulas

A reduction formula expresses $\int f^n\,dx$ in terms of $\int f^{n-2}\,dx$ or $\int f^{n-1}\,dx$ . Applied repeatedly it grinds any power down to a base case, and it is derived once, by parts, rather than rediscovered every time. Two derivations are given in full; the others follow the same pattern.

Theorem 5.9 (Reduction formula for powers of sine). For every integer $n \ge 2$ ,

\int \sin^n x\,dx = -\frac{1}{n}\sin^{n-1}x\cos x + \frac{n-1}{n}\int \sin^{n-2}x\,dx.

Proof. Write $\sin^n x = \sin^{n-1}x\cdot\sin x$ and integrate by parts with $u = \sin^{n-1}x$ , $dv = \sin x\,dx$ , so that $du = (n-1)\sin^{n-2}x\cos x\,dx$ and $v = -\cos x$ :

\int \sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int \sin^{n-2}x\cos^2 x\,dx.

Replace $\cos^{2} x$ by $1 - \sin^{2} x$ in the last integral. It splits into $(n-1)\int\sin^{n-2}x\,dx - (n-1)\int\sin^n x\,dx$ , and the second piece is $(n-1)$ times the integral we started with. Moving it to the left,

n\int\sin^n x\,dx = -\sin^{n-1}x\cos x + (n-1)\int\sin^{n-2}x\,dx,

and dividing by $n$ gives the formula.∎

Theorem 5.10 (Reduction formula for powers of secant). For every integer $n \ge 2$ ,

\int \sec^n x\,dx = \frac{\sec^{n-2}x\tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2}x\,dx.

Proof. Write $\sec^n x = \sec^{n-2}x\cdot\sec^{2} x$ and take $u = \sec^{n-2}x$ , $dv = \sec^{2} x\,dx$ , so $du = (n-2)\sec^{n-2}x\tan x\,dx$ and $v = \tan x$ :

\int\sec^n x\,dx = \sec^{n-2}x\tan x - (n-2)\int\sec^{n-2}x\tan^2 x\,dx.

Replace $\tan^{2} x$ by $\sec^{2} x - 1$ . The integral becomes $(n-2)\int\sec^n x\,dx - (n-2)\int\sec^{n-2}x\,dx$ , so

(n-1)\int\sec^n x\,dx = \sec^{n-2}x\tan x + (n-2)\int\sec^{n-2}x\,dx.

Divide by $n-1$ .∎

The pattern in both proofs is the same: peel off one or two factors to serve as $dv$ , differentiate what is left, use a Pythagorean identity to express the new integrand in terms of the original power and a lower power, and solve for the original integral. The remaining standard formulas come from the same recipe.

Proposition 5.11 (Further reduction formulas). For integers $n \ge 2$ (for $x^n e^x$ and $(\ln x)^n$ , $n \ge 1$ ):

\int\cos^n x\,dx = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}\int\cos^{n-2}x\,dx,

\int\tan^n x\,dx = \frac{\tan^{n-1}x}{n-1} - \int\tan^{n-2}x\,dx,

\int x^n e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx,

\int(\ln x)^n\,dx = x(\ln x)^n - n\int(\ln x)^{n-1}\,dx.

The cosine formula is the sine proof with the roles of $\sin$ and $\cos$ exchanged. The tangent formula needs no parts at all: write $\tan^n x = \tan^{n-2}x(\sec^{2} x - 1)$ and substitute $u = \tan x$ in the first piece. The last two are single applications of parts with $u = x^n$ or $u = (\ln x)^n$ and $dv = e^x\,dx$ or $dv = dx$ .

Intuition. A reduction formula is a staircase. To compute $\int\sin^{6} x\,dx$ you do not need six flights of cleverness, only the same step three times: $\sin^{6}$ leads to $\sin^{4}$ , which leads to $\sin^{2}$ , which leads to $\sin^{0} = 1$ . Each step costs one term of the form $\sin^{k-1}x\cos x$ and one rational factor. For the definite integral over $[0, \pi/2]$ even those terms vanish, because $\sin 0 = 0$ and $\cos(\pi/2) = 0$ , and the whole computation is a product of fractions.

Example 5.12 (Wallis integrals). Let $I_n = \int_0^{\pi/2}\sin^n x\,dx$ . Show that $I_n = \frac{n-1}{n}I_{n-2}$ for $n \ge 2$ and compute $I_4$ .

Solution.

Apply the sine reduction formula between $0$ and $\pi/2$ . The boundary term is

\Big[-\frac1n\sin^{n-1}x\cos x\Big]_0^{\pi/2} = -\frac1n\big(\sin^{n-1}\tfrac{\pi}{2}\cos\tfrac{\pi}{2} - \sin^{n-1}0\cos 0\big) = -\frac1n(1\cdot 0 - 0\cdot 1) = 0,

so $I_n = \frac{n-1}{n}I_{n-2}$ .

The base cases are $I_0 = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}$ and $I_1 = \int_0^{\pi/2}\sin x\,dx = 1$ .
Then $I_2 = \frac12 I_0 = \frac{\pi}{4}$ and $I_4 = \frac34 I_2 = \frac{3\pi}{16}$ .
Sanity check: $0 \le \sin^{4} x \le \sin^{2} x$ on $[0, \pi/2]$ , so $I_4$ should be smaller than $I_2 = \pi/4 \approx 0.785$ . Indeed $3\pi/16 \approx 0.589$ .

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In general $I_{2k} = \frac{(2k-1)(2k-3)\cdots 1}{(2k)(2k-2)\cdots 2}\cdot\frac{\pi}{2}$ and $I_{2k+1} = \frac{(2k)(2k-2)\cdots 2}{(2k+1)(2k-1)\cdots 3}$ . These are the Wallis integrals; the ratio $I_{2k}/I_{2k+1} \to 1$ gives Wallis's infinite product for $\pi$ .

Example 5.13 (A power of the logarithm). Compute $\int(\ln x)^2\,dx$ .

Solution.

The reduction formula with $n = 2$ gives $\int(\ln x)^2\,dx = x(\ln x)^2 - 2\int\ln x\,dx$ .
With $n = 1$ it gives $\int\ln x\,dx = x\ln x - \int 1\,dx = x\ln x - x$ .
Combining, $\int(\ln x)^2\,dx = x(\ln x)^2 - 2x\ln x + 2x + C$ .
Check: the derivative of $x(\ln x)^2$ is $(\ln x)^2 + 2\ln x$ ; of $-2x\ln x$ is $-2\ln x - 2$ ; of $2x$ is $2$ . The sum is $(\ln x)^2$ .

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Example 5.14 (A power of the tangent). Compute $\int\tan^{4} x\,dx$ .

Solution.

With $n = 4$ : $\int\tan^{4} x\,dx = \frac{\tan^{3} x}{3} - \int\tan^{2} x\,dx$ .
With $n = 2$ : $\int\tan^{2} x\,dx = \tan x - \int 1\,dx = \tan x - x$ .
So $\int\tan^{4} x\,dx = \frac{\tan^{3} x}{3} - \tan x + x + C$ .
Check: $\frac{d}{dx}\big(\tfrac13\tan^{3} x\big) = \tan^{2} x\sec^{2} x$ , and $\sec^{2} x = 1 + \tan^{2} x$ , so this is $\tan^{2} x + \tan^{4} x$ . Subtracting $\sec^{2} x = 1 + \tan^{2} x$ and adding $1$ leaves $\tan^{4} x$ .

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Pitfall. Each formula has a floor. The sine, cosine, secant and tangent formulas all need $n \ge 2$ ; the base cases $\int\sin x$ , $\int\sec x$ , $\int\tan x$ and $\int 1$ must be known separately. In a definite integral the boundary term $\big[\cdot\big]_a^b$ must be evaluated at every step, not only at the end; it happened to vanish in the Wallis example because of the specific limits $0$ and $\pi/2$ .

Exercises for §5.2 →

5.3Trigonometric integrals

Products of powers of trigonometric functions integrate by a fixed playbook: use an identity to arrange the integrand as (a function of $u$ ) times $du$ , then substitute. Which identity, and which $u$ , is decided by the parities of the exponents. The three identities that do all the work are

\sin^2 x + \cos^2 x = 1, \qquad 1 + \tan^2 x = \sec^2 x, \qquad \sin^2 x = \frac{1 - \cos 2x}{2},\quad \cos^2 x = \frac{1 + \cos 2x}{2}.

Method 5.15 (Integrals of $\sin^m x\cos^n x$ ).

If $n$ is odd, save one factor of $\cos x$ for $du$ , convert the remaining even power with $\cos^{2} x = 1 - \sin^{2} x$ , and substitute $u = \sin x$ .
If $m$ is odd, save one factor of $\sin x$ , convert the rest with $\sin^{2} x = 1 - \cos^{2} x$ , and substitute $u = \cos x$ (remember $du = -\sin x\,dx$ ).
If both are odd, either route works.
If both are even, use the half-angle identities to halve the degree, expand, and repeat on any remaining even powers. Products such as $\sin x\cos x$ become $\tfrac12\sin 2x$ .

Method 5.16 (Integrals of $\tan^m x\sec^n x$ ).

If $n$ is even and $n \ge 2$ , save one factor of $\sec^{2} x$ for $du$ , convert the remaining $\sec^{n-2}x$ with $\sec^{2} x = 1 + \tan^{2} x$ , and substitute $u = \tan x$ .
If $m$ is odd, save one factor of $\sec x\tan x$ for $du$ , convert the remaining $\tan^{m-1}x$ with $\tan^{2} x = \sec^{2} x - 1$ , and substitute $u = \sec x$ .
If $m$ is even and $n$ is odd, no substitution fits. Convert all the tangents with $\tan^{2} x = \sec^{2} x - 1$ and integrate the resulting powers of $\sec x$ with the secant reduction formula, whose base cases are $\int\sec x\,dx$ and $\int\sec^{3} x\,dx$ below.

The reason the odd cases are easy is that an odd power always has one factor to spare after the rest have been paired into squares, and that spare factor is exactly the derivative of the other function. The even case has no factor to spare, which is why it needs the half-angle identities instead.

Proposition 5.17 (The four basic trigonometric integrals).

\int\tan x\,dx = \ln|\sec x| + C, \qquad \int\cot x\,dx = \ln|\sin x| + C,

\int\sec x\,dx = \ln|\sec x + \tan x| + C, \qquad \int\csc x\,dx = \ln|\csc x - \cot x| + C.

Proof. For the first, write $\tan x = \frac{\sin x}{\cos x}$ and substitute $u = \cos x$ , $du = -\sin x\,dx$ : the integral is $-\int\frac{du}{u} = -\ln|\cos x| + C = \ln|\sec x| + C$ . The cotangent is the same with $u = \sin x$ .

For the secant, multiply and divide by $\sec x + \tan x$ :

\int\sec x\,dx = \int\frac{\sec^2 x + \sec x\tan x}{\sec x + \tan x}\,dx.

The numerator is exactly the derivative of the denominator, since $\frac{d}{dx}(\sec x + \tan x) = \sec x\tan x + \sec^{2} x$ . With $u = \sec x + \tan x$ the integral is $\int\frac{du}{u} = \ln|u| + C$ . The cosecant formula comes from the same trick with $\csc x - \cot x$ , whose derivative is $-\csc x\cot x + \csc^{2} x$ .∎

The secant formula is famous for being unguessable; the multiplication by $\sec x + \tan x$ looks like a conjuring trick. It is worth knowing that the result can also be reached honestly, by writing $\sec x = \frac{\cos x}{\cos^{2} x} = \frac{\cos x}{1 - \sin^{2} x}$ , substituting $u = \sin x$ and using partial fractions on $\frac{1}{1-u^{2}}$ ; that route gives $\tfrac12\ln\left|\frac{1+\sin x}{1-\sin x}\right|$ , which is the same function.

Proposition 5.18 (The integral of secant cubed).

\int\sec^3 x\,dx = \frac12\sec x\tan x + \frac12\ln|\sec x + \tan x| + C.

Proof. Integrate by parts with $u = \sec x$ , $dv = \sec^{2} x\,dx$ , so $du = \sec x\tan x\,dx$ and $v = \tan x$ :

\int\sec^3 x\,dx = \sec x\tan x - \int\sec x\tan^2 x\,dx = \sec x\tan x - \int\sec x(\sec^2 x - 1)\,dx.

The last integral is $\int\sec^{3} x\,dx - \int\sec x\,dx$ . Writing $I$ for $\int\sec^{3} x\,dx$ this is the boomerang equation $I = \sec x\tan x - I + \ln|\sec x + \tan x|$ , so $2I = \sec x\tan x + \ln|\sec x + \tan x|$ .∎

This is the $n = 3$ case of the secant reduction formula, and it appears whenever a trigonometric substitution meets $\sqrt{x^{2} + a^{2}}$ , so it is worth knowing by heart.

Proposition 5.19 (Product-to-sum identities).

\sin A\cos B = \tfrac12\big[\sin(A - B) + \sin(A + B)\big],

\sin A\sin B = \tfrac12\big[\cos(A - B) - \cos(A + B)\big],

\cos A\cos B = \tfrac12\big[\cos(A - B) + \cos(A + B)\big].

Proof. Add and subtract the angle-sum formulas. For instance $\sin(A+B) = \sin A\cos B + \cos A\sin B$ and $\sin(A-B) = \sin A\cos B - \cos A\sin B$ ; adding them and halving gives the first identity. The other two come from the two cosine formulas $\cos(A \mp B) = \cos A\cos B \pm \sin A\sin B$ .∎

These identities turn a product with different frequencies, such as $\sin 4x\cos 2x$ , into a sum of single sines and cosines that integrate term by term. They are also the reason the functions $\sin nx$ and $\cos nx$ are orthogonal on $[-\pi, \pi]$ , the fact underlying Fourier series.

Intuition. Think of an odd power as having a spare part. In $\sin^{5} x\cos^{2} x$ , four of the five sines pair up into $(\sin^{2} x)^2$ , which the identity rewrites in terms of $\cos x$ ; the fifth sine is left over, and it is exactly what $du = -\sin x\,dx$ needs. With $\sin^{4} x\cos^{2} x$ there is no spare part: every factor is locked in a square, so the substitution has nothing to grab, and you must use half-angles to lower the degree by brute force.

Example 5.20 (An odd power of sine). Compute $\int\sin^{5} x\cos^{2} x\,dx$ .

Solution.

The power of $\sin x$ is odd. Save one factor: $\sin^{5} x\cos^{2} x = (\sin^{2} x)^2\cos^{2} x\sin x = (1 - \cos^{2} x)^2\cos^{2} x\sin x$ .
Substitute $u = \cos x$ , $du = -\sin x\,dx$ :

\int(1 - u^2)^2u^2(-du) = -\int(u^2 - 2u^4 + u^6)\,du = -\frac{u^3}{3} + \frac{2u^5}{5} - \frac{u^7}{7} + C.

Back in $x$ : $\int\sin^{5} x\cos^{2} x\,dx = -\frac{\cos^{3} x}{3} + \frac{2\cos^{5} x}{5} - \frac{\cos^{7} x}{7} + C$ .
Check the first term: $\frac{d}{dx}\big(-\tfrac13\cos^{3} x\big) = \cos^{2} x\sin x$ , and the pattern continues; the derivative of the whole is $\sin x\cos^{2} x(1 - 2\cos^{2} x + \cos^{4} x) = \sin x\cos^{2} x(1 - \cos^{2} x)^2 = \sin^{5} x\cos^{2} x$ .

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Example 5.21 (Both powers even). Evaluate $\int_0^{\pi}\sin^{2} x\cos^{2} x\,dx$ .

Solution.

Both exponents are even, so no substitution is available. First shrink the product: $\sin x\cos x = \tfrac12\sin 2x$ , so $\sin^{2} x\cos^{2} x = \tfrac14\sin^{2} 2x$ .
Half-angle again: $\sin^{2} 2x = \frac{1 - \cos 4x}{2}$ , so the integrand is $\frac{1 - \cos 4x}{8}$ .
Integrate:

\int_0^{\pi}\frac{1 - \cos 4x}{8}\,dx = \Big[\frac{x}{8} - \frac{\sin 4x}{32}\Big]_0^{\pi} = \frac{\pi}{8} - 0 = \frac{\pi}{8}.

Sanity check: $\sin^{2} 2x$ averages $\tfrac12$ over a whole number of periods, so the integrand averages $\tfrac18$ over $[0, \pi]$ , and $\tfrac18\cdot\pi$ is right.

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Example 5.22 (Tangent and secant, two routes). Compute $\int\tan^{3} x\sec^{4} x\,dx$ .

Solution.

The power of $\sec x$ is even, so save $\sec^{2} x$ and convert the rest: $\tan^{3} x\sec^{4} x = \tan^{3} x(1 + \tan^{2} x)\sec^{2} x$ . With $u = \tan x$ , $du = \sec^{2} x\,dx$ :

\int(u^3 + u^5)\,du = \frac{u^4}{4} + \frac{u^6}{6} + C = \frac{\tan^4 x}{4} + \frac{\tan^6 x}{6} + C.

The power of $\tan x$ is also odd, so a second route exists. Save $\sec x\tan x$ : $\tan^{3} x\sec^{4} x = \tan^{2} x\sec^{3} x\cdot\sec x\tan x = (\sec^{2} x - 1)\sec^{3} x\cdot\sec x\tan x$ . With $u = \sec x$ , $du = \sec x\tan x\,dx$ :

\int(u^5 - u^3)\,du = \frac{\sec^6 x}{6} - \frac{\sec^4 x}{4} + C.

The two answers look different but differ by a constant. Writing $s = \sec^{2} x = 1 + t^{2}$ with $t = \tan x$ , the second is $\frac{(1+t^{2})^3}{6} - \frac{(1+t^{2})^2}{4} = \frac{t^{6}}{6} + \frac{t^{4}}{4} - \frac{1}{12}$ , which is the first answer minus $\tfrac{1}{12}$ . Both are correct.

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Example 5.23 (Even tangent, odd secant). Compute $\int\tan^{2} x\sec x\,dx$ .

Solution.

Neither substitution applies: saving $\sec^{2} x$ leaves $\sec^{-1}x$ , and saving $\sec x\tan x$ leaves $\tan x$ , an odd leftover the identity cannot convert.
Convert the tangent instead: $\tan^{2} x\sec x = (\sec^{2} x - 1)\sec x = \sec^{3} x - \sec x$ .
Both pieces are known:

\int\tan^2 x\sec x\,dx = \Big(\frac12\sec x\tan x + \frac12\ln|\sec x + \tan x|\Big) - \ln|\sec x + \tan x| + C = \frac12\sec x\tan x - \frac12\ln|\sec x + \tan x| + C.

Check: differentiate $\tfrac12\sec x\tan x$ to get $\tfrac12(\sec x\tan^{2} x + \sec^{3} x)$ , and differentiate $-\tfrac12\ln|\sec x + \tan x|$ to get $-\tfrac12\sec x$ . Since $\sec^{3} x - \sec x = \sec x\tan^{2} x$ , the total is $\tfrac12\sec x\tan^{2} x + \tfrac12\sec x\tan^{2} x = \tan^{2} x\sec x$ .

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Example 5.24 (Different frequencies). Compute $\int\sin 4x\cos 2x\,dx$ .

Solution.

This is a product of a sine and a cosine with different arguments; no power trick applies. Use the product-to-sum identity with $A = 4x$ , $B = 2x$ :

\sin 4x\cos 2x = \tfrac12\big[\sin 2x + \sin 6x\big].

Integrate term by term:

\int\sin 4x\cos 2x\,dx = \frac12\Big(-\frac{\cos 2x}{2} - \frac{\cos 6x}{6}\Big) + C = -\frac{\cos 2x}{4} - \frac{\cos 6x}{12} + C.

Check: the derivative is $\tfrac12\sin 2x + \tfrac12\sin 6x$ , which the identity says equals $\sin 4x\cos 2x$ .

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Pitfall. $\int\sin^{2} x\,dx$ is not $\frac{\sin^{3} x}{3}$ . The power rule applies to $u^n\,du$ , and $\sin^{2} x\,dx$ has no $du = \cos x\,dx$ attached. The even case always goes through the half-angle identity: $\int\sin^{2} x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$ . The same error in the other direction is trying to use half-angles on $\sin^{3} x$ , which works but is far longer than saving one sine and substituting.

Exercises for §5.3 →

5.4Trigonometric substitution

No algebraic manipulation removes the square root from $\sqrt{a^{2} - x^{2}}$ , $\sqrt{a^{2} + x^{2}}$ or $\sqrt{x^{2} - a^{2}}$ . What does remove it is the substitution rule run backwards: instead of replacing an expression in $x$ by a new variable $u$ , replace $x$ itself by a trigonometric function of a new angle $\theta$ , chosen so that a Pythagorean identity turns the radicand into a perfect square.

Method 5.25 (Trigonometric substitution).

For $\sqrt{a^{2} - x^{2}}$ : put $x = a\sin\theta$ with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ . Then $dx = a\cos\theta\,d\theta$ and $\sqrt{a^{2} - x^{2}} = a\cos\theta$ .
For $\sqrt{a^{2} + x^{2}}$ : put $x = a\tan\theta$ with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ . Then $dx = a\sec^{2}\theta\,d\theta$ and $\sqrt{a^{2} + x^{2}} = a\sec\theta$ .
For $\sqrt{x^{2} - a^{2}}$ : put $x = a\sec\theta$ with $0 \le \theta < \frac{\pi}{2}$ when $x \ge a$ (and $\pi \le \theta < \frac{3\pi}{2}$ when $x \le -a$ ). Then $dx = a\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^{2} - a^{2}} = a\tan\theta$ .
Substitute for $x$ , $dx$ and the root, integrate the resulting trigonometric integral, and convert back to $x$ with a reference triangle. For a definite integral, convert the limits to $\theta$ instead and never return to $x$ .
If the quadratic under the root is not in one of the three forms, complete the square first and substitute $u = x + \text{const}$ to bring it into one.

Proposition 5.26 (Why the root disappears). With $x = a\sin\theta$ and $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ ,

\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2\cos^2\theta} = a|\cos\theta| = a\cos\theta,

because $\cos\theta \ge 0$ on that interval. The other two cases are identical with $1 + \tan^{2}\theta = \sec^{2}\theta$ and $\sec^{2}\theta - 1 = \tan^{2}\theta$ ; the stated ranges of $\theta$ are exactly those on which $\sec\theta > 0$ , respectively $\tan\theta \ge 0$ , so that the absolute value can be dropped.

The range of $\theta$ is not decoration. It makes the substitution one-to-one, so that $\theta$ can be recovered from $x$ as $\arcsin(x/a)$ , $\arctan(x/a)$ or, for $x \ge a$ , $\operatorname{arcsec}(x/a)$ , and it fixes the sign of the root. Without it $\sqrt{a^{2}\cos^{2}\theta}$ would be $a|\cos\theta|$ and every subsequent line would need a case split.

The reference triangle is the tool for converting back. For $x = a\sin\theta$ , draw a right triangle with an acute angle $\theta$ , label the opposite side $x$ and the hypotenuse $a$ , so that $\sin\theta = x/a$ ; the Pythagorean theorem gives the adjacent side $\sqrt{a^{2} - x^{2}}$ , and every trigonometric function of $\theta$ can now be read off. For $x = a\tan\theta$ the opposite side is $x$ and the adjacent side is $a$ , with hypotenuse $\sqrt{x^{2} + a^{2}}$ . For $x = a\sec\theta$ the hypotenuse is $x$ and the adjacent side is $a$ , with opposite side $\sqrt{x^{2} - a^{2}}$ .

Intuition. A point $(x, y)$ on the circle $x^{2} + y^{2} = 9$ has $y = \sqrt{9 - x^{2}}$ , and that square root is awkward to work with. But the same point is $(3\cos\phi, 3\sin\phi)$ , and there is no square root anywhere: the circle's own coordinate system does the algebra for you. Trigonometric substitution is nothing more than switching to that coordinate system. The substitution $x = 3\sin\theta$ says: measure position on the circle by an angle, and the root $\sqrt{9 - x^{2}}$ becomes the plain coordinate $3\cos\theta$ .

Example 5.27 (The $\sqrt{a^{2} - x^{2}}$ pattern with a reference triangle). Compute $\int\frac{\sqrt{9 - x^{2}}}{x^{2}}\,dx$ .

Solution.

The root is $\sqrt{a^{2} - x^{2}}$ with $a = 3$ . Put $x = 3\sin\theta$ , $dx = 3\cos\theta\,d\theta$ , $\sqrt{9 - x^{2}} = 3\cos\theta$ .
Substitute everything:

\int\frac{3\cos\theta}{9\sin^2\theta}\cdot 3\cos\theta\,d\theta = \int\frac{\cos^2\theta}{\sin^2\theta}\,d\theta = \int\cot^2\theta\,d\theta.

Use $\cot^{2}\theta = \csc^{2}\theta - 1$ : the integral is $-\cot\theta - \theta + C$ .
Convert back. From $\sin\theta = x/3$ , the reference triangle has opposite side $x$ , hypotenuse $3$ and adjacent side $\sqrt{9 - x^{2}}$ , so $\cot\theta = \frac{\sqrt{9 - x^{2}}}{x}$ ; and $\theta = \arcsin\frac{x}{3}$ . Hence

\int\frac{\sqrt{9 - x^2}}{x^2}\,dx = -\frac{\sqrt{9 - x^2}}{x} - \arcsin\frac{x}{3} + C.

Check: differentiating $-\frac{\sqrt{9-x^{2}}}{x}$ by the quotient rule gives $\frac{9}{x^{2}\sqrt{9 - x^{2}}}$ , and differentiating $-\arcsin\frac{x}{3}$ gives $-\frac{1}{\sqrt{9 - x^{2}}} = -\frac{x^{2}}{x^{2}\sqrt{9-x^{2}}}$ . The sum is $\frac{9 - x^{2}}{x^{2}\sqrt{9 - x^{2}}} = \frac{\sqrt{9 - x^{2}}}{x^{2}}$ .

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Example 5.28 (The $\sqrt{a^{2} + x^{2}}$ pattern). Compute $\int\frac{dx}{x^{2}\sqrt{x^{2} + 4}}$ .

Solution.

Put $x = 2\tan\theta$ , $dx = 2\sec^{2}\theta\,d\theta$ , $\sqrt{x^{2} + 4} = 2\sec\theta$ .
Substitute:

\int\frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot 2\sec\theta} = \frac14\int\frac{\sec\theta}{\tan^2\theta}\,d\theta = \frac14\int\frac{\cos\theta}{\sin^2\theta}\,d\theta.

With $u = \sin\theta$ , $du = \cos\theta\,d\theta$ , this is $\frac14\int u^{-2}\,du = -\frac{1}{4u} = -\frac{1}{4}\csc\theta + C$ .
Reference triangle for $\tan\theta = x/2$ : opposite $x$ , adjacent $2$ , hypotenuse $\sqrt{x^{2} + 4}$ , so $\csc\theta = \frac{\sqrt{x^{2} + 4}}{x}$ . Therefore

\int\frac{dx}{x^2\sqrt{x^2 + 4}} = -\frac{\sqrt{x^2 + 4}}{4x} + C.

Sanity check at large $x$ : the integrand behaves like $x^{-3}$ , whose antiderivative behaves like $-\tfrac12 x^{-2}$ ; and $-\frac{\sqrt{x^{2}+4}}{4x} \approx -\frac{1}{4}$ , a constant. Two antiderivatives may differ by a constant, so this is consistent only if $-\frac{\sqrt{x^{2}+4}}{4x} + \frac14 \approx -\tfrac12 x^{-2}$ , and indeed $\sqrt{x^{2} + 4} = x\sqrt{1 + 4/x^{2}} \approx x + 2/x$ gives $-\frac{\sqrt{x^{2}+4}}{4x} \approx -\frac14 - \frac{1}{2x^{2}}$ .

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Example 5.29 (The $\sqrt{x^{2} - a^{2}}$ pattern and a logarithm). Show that for $x > a > 0$ ,

\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C.

Solution.

Put $x = a\sec\theta$ with $0 \le \theta < \frac{\pi}{2}$ , so $dx = a\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^{2} - a^{2}} = a\tan\theta$ .
Substitute:

\int\frac{a\sec\theta\tan\theta}{a\tan\theta}\,d\theta = \int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C.

From the reference triangle with hypotenuse $x$ and adjacent side $a$ , $\sec\theta = \frac{x}{a}$ and $\tan\theta = \frac{\sqrt{x^{2} - a^{2}}}{a}$ , so

\ln\Big|\frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a}\Big| + C = \ln\left|x + \sqrt{x^2 - a^2}\right| - \ln a + C.

The term $-\ln a$ is a constant and is absorbed into $C$ .
Check: with $F = \ln(x + \sqrt{x^{2} - a^{2}})$ , $F' = \frac{1 + x/\sqrt{x^{2} - a^{2}}}{x + \sqrt{x^{2} - a^{2}}} = \frac{(\sqrt{x^{2} - a^{2}} + x)/\sqrt{x^{2} - a^{2}}}{x + \sqrt{x^{2} - a^{2}}} = \frac{1}{\sqrt{x^{2} - a^{2}}}$ .

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The companion formula $\int\frac{dx}{\sqrt{x^{2} + a^{2}}} = \ln\left|x + \sqrt{x^{2} + a^{2}}\right| + C$ comes from $x = a\tan\theta$ in the same way. Both are worth memorising, as is

\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C,

which is $x = a\sin\theta$ followed by $a^{2}\int\cos^{2}\theta\,d\theta = \frac{a^{2}}{2}(\theta + \sin\theta\cos\theta)$ .

Example 5.30 (Completing the square first). Compute $\int\frac{x}{\sqrt{3 - 2x - x^{2}}}\,dx$ .

Solution.

The radicand is not in standard form. Complete the square: $3 - 2x - x^{2} = 3 - (x^{2} + 2x) = 3 - \big((x + 1)^2 - 1\big) = 4 - (x + 1)^2$ .
Substitute $u = x + 1$ , so $x = u - 1$ and $dx = du$ :

\int\frac{u - 1}{\sqrt{4 - u^2}}\,du.

Now put $u = 2\sin\theta$ , $du = 2\cos\theta\,d\theta$ , $\sqrt{4 - u^{2}} = 2\cos\theta$ :

\int\frac{2\sin\theta - 1}{2\cos\theta}\cdot 2\cos\theta\,d\theta = \int(2\sin\theta - 1)\,d\theta = -2\cos\theta - \theta + C.

Convert back: $\cos\theta = \frac{\sqrt{4 - u^{2}}}{2}$ and $\theta = \arcsin\frac{u}{2}$ , so the integral is $-\sqrt{4 - u^{2}} - \arcsin\frac{u}{2} + C$ , that is,

\int\frac{x}{\sqrt{3 - 2x - x^2}}\,dx = -\sqrt{3 - 2x - x^2} - \arcsin\frac{x + 1}{2} + C.

Check: the derivative of $-\sqrt{3 - 2x - x^{2}}$ is $\frac{2 + 2x}{2\sqrt{3 - 2x - x^{2}}} = \frac{x + 1}{\sqrt{3 - 2x - x^{2}}}$ , and the derivative of $-\arcsin\frac{x+1}{2}$ is $-\frac{1}{\sqrt{4 - (x+1)^2}} = -\frac{1}{\sqrt{3 - 2x - x^{2}}}$ . The sum is $\frac{x}{\sqrt{3 - 2x - x^{2}}}$ .

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Example 5.31 (A definite integral with converted limits). Evaluate $\int_0^3\sqrt{9 - x^{2}}\,dx$ .

Solution.

Put $x = 3\sin\theta$ , $dx = 3\cos\theta\,d\theta$ , $\sqrt{9 - x^{2}} = 3\cos\theta$ .
Convert the limits: $x = 0$ gives $\sin\theta = 0$ , so $\theta = 0$ ; $x = 3$ gives $\sin\theta = 1$ , so $\theta = \frac{\pi}{2}$ . Both lie in the allowed range.
The integral becomes

\int_0^{\pi/2}3\cos\theta\cdot 3\cos\theta\,d\theta = 9\int_0^{\pi/2}\cos^2\theta\,d\theta = 9\Big[\frac{\theta}{2} + \frac{\sin 2\theta}{4}\Big]_0^{\pi/2} = 9\cdot\frac{\pi}{4} = \frac{9\pi}{4}.

Sanity check: the graph of $y = \sqrt{9 - x^{2}}$ on $[0, 3]$ is a quarter of the circle of radius $3$ , whose area is $\frac14\pi\cdot 9 = \frac{9\pi}{4}$ .

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Pitfall. Three things are forgotten most often. The differential: $dx$ becomes $a\cos\theta\,d\theta$ , not $d\theta$ . The limits: in a definite integral they must be converted to $\theta$ or the answer must be converted back to $x$ before the limits are used, never a mixture. And the sign for $x \le -a$ in the secant case: with the range $\pi \le \theta < \frac{3\pi}{2}$ the identity $\sqrt{x^{2} - a^{2}} = a\tan\theta$ still holds, but the reference triangle must be read with care; the safest route is to substitute $x = -t$ and reduce to the case $t \ge a$ .

Exercises for §5.4 →

5.5Partial fractions

A rational function is a quotient of polynomials, $\frac{P(x)}{Q(x)}$ . Integrating one is an algebra problem followed by a short list of standard integrals: break the fraction into a sum of simpler fractions, each with a denominator that is a power of a linear or an irreducible quadratic factor of $Q$ , and integrate the pieces separately. The algebra is called partial fraction decomposition.

Definition 5.32 (Proper rational function). The rational function $\frac{P(x)}{Q(x)}$ is proper if $\deg P < \deg Q$ . If $\deg P \ge \deg Q$ , polynomial long division writes

\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)},

with $S$ a polynomial and $\deg R < \deg Q$ , so that the remainder fraction is proper.

Theorem 5.33 (Partial fraction decomposition). Every real polynomial $Q$ factors into linear factors $(ax + b)$ and irreducible quadratic factors $(ax^{2} + bx + c)$ with $b^{2} - 4ac < 0$ . If $\frac{R}{Q}$ is proper, it can be written uniquely as a sum of terms determined by the factors of $Q$ :

Each non-repeated linear factor $(ax + b)$ contributes $\dfrac{A}{ax + b}$ .
Each repeated linear factor $(ax + b)^k$ contributes $\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax + b)^2} + \cdots + \dfrac{A_k}{(ax + b)^k}$ .
Each non-repeated irreducible quadratic factor contributes $\dfrac{Ax + B}{ax^{2} + bx + c}$ .
Each repeated irreducible quadratic factor $(ax^{2} + bx + c)^k$ contributes $\dfrac{A_1x + B_1}{ax^{2} + bx + c} + \cdots + \dfrac{A_kx + B_k}{(ax^{2} + bx + c)^k}$ .

The factorisation statement is the real form of the fundamental theorem of algebra, and the existence and uniqueness of the decomposition is a theorem of algebra whose proof does not belong to calculus. What calculus needs is the shape of the answer, and the theorem gives exactly that: it tells you how many unknown constants to write down and where. The numerator over a quadratic factor is linear, $Ax + B$ , because a proper fraction over a quadratic has degree at most one on top; the numerator over a linear factor is a constant for the same reason.

Proposition 5.34 (The cover-up method). If $Q(x) = (x - r)\,Q_1(x)$ with $Q_1(r) \ne 0$ , then in the decomposition of $\frac{R(x)}{Q(x)}$ the coefficient $A$ of $\frac{1}{x - r}$ is

A = \frac{R(r)}{Q_1(r)},

that is, the value at $x = r$ of the original fraction with the factor $(x - r)$ covered up.

Proof. Multiply the decomposition $\frac{R(x)}{(x - r)Q_1(x)} = \frac{A}{x - r} + (\text{other terms})$ by $(x - r)$ . The left side becomes $\frac{R(x)}{Q_1(x)}$ , the first term becomes $A$ , and every other term still carries a factor $(x - r)$ in its numerator, because their denominators do not contain $(x - r)$ . Now let $x \to r$ : the other terms vanish and $A = \frac{R(r)}{Q_1(r)}$ .∎

The cover-up method produces every coefficient over a non-repeated linear factor instantly, and the top coefficient $A_k$ over a repeated factor $(x - r)^k$ by the same argument (cover up $(x - r)^k$ ). The remaining coefficients are found by clearing denominators and comparing coefficients of powers of $x$ , or by substituting further convenient values of $x$ .

Method 5.35 (Integrating a rational function).

If the fraction is improper, divide first. The polynomial part integrates by the power rule.
Factor the denominator completely into linear and irreducible quadratic factors.
Write the decomposition with unknown coefficients, following the four cases of the theorem.
Find the coefficients: cover-up for the simple linear ones, then clear denominators and compare coefficients or substitute values for the rest.
Integrate each piece: $\int\frac{A}{x - r}\,dx = A\ln|x - r|$ ; $\int\frac{A}{(x - r)^k}\,dx = \frac{A}{(1 - k)(x - r)^{k-1}}$ for $k \ge 2$ ; and for a quadratic piece complete the square, split the numerator into a multiple of the derivative of the denominator plus a constant, and use $\int\frac{du}{u} = \ln|u|$ together with $\int\frac{du}{u^{2} + a^{2}} = \frac1a\arctan\frac{u}{a}$ .

The quadratic piece deserves one worked line. To integrate $\frac{2x + 5}{x^{2} + 2x + 5}$ , note that the derivative of the denominator is $2x + 2$ , so $2x + 5 = (2x + 2) + 3$ and

\int\frac{2x + 5}{x^2 + 2x + 5}\,dx = \int\frac{2x + 2}{x^2 + 2x + 5}\,dx + \int\frac{3}{(x + 1)^2 + 4}\,dx = \ln(x^2 + 2x + 5) + \frac32\arctan\frac{x + 1}{2} + C.

The logarithm needs no absolute value because an irreducible quadratic is never zero, hence always positive.

Intuition. Adding fractions is easy: $\frac{2}{x - 1} - \frac{1}{x + 2} = \frac{2(x + 2) - (x - 1)}{(x - 1)(x + 2)} = \frac{x + 5}{(x - 1)(x + 2)}$ . Partial fractions is that computation run in reverse. Somebody has handed you the combined fraction, and you are recovering the simple pieces it was built from, because each simple piece is something you know how to integrate and the combined fraction is not.

Example 5.36 (Distinct linear factors). Compute $\int\frac{x + 5}{x^{2} + x - 2}\,dx$ .

Solution.

The fraction is proper ( $1 < 2$ ). Factor the denominator: $x^{2} + x - 2 = (x - 1)(x + 2)$ .
Write $\frac{x + 5}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ .
Cover-up. For $A$ , cover $(x - 1)$ and set $x = 1$ : $A = \frac{1 + 5}{1 + 2} = 2$ . For $B$ , cover $(x + 2)$ and set $x = -2$ : $B = \frac{-2 + 5}{-2 - 1} = -1$ .
Integrate:

\int\frac{x + 5}{x^2 + x - 2}\,dx = 2\ln|x - 1| - \ln|x + 2| + C.

Check by recombining: $\frac{2}{x - 1} - \frac{1}{x + 2} = \frac{2x + 4 - x + 1}{(x - 1)(x + 2)} = \frac{x + 5}{(x - 1)(x + 2)}$ .

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Example 5.37 (An improper fraction: divide first). Compute $\int\frac{x^{3} + x}{x - 1}\,dx$ .

Solution.

The numerator has degree $3$ and the denominator degree $1$ , so divide. Long division gives $x^{3} + x = (x - 1)(x^{2} + x + 2) + 2$ , so

\frac{x^3 + x}{x - 1} = x^2 + x + 2 + \frac{2}{x - 1}.

The remainder fraction has a linear denominator and needs no further decomposition.
Integrate term by term:

\int\frac{x^3 + x}{x - 1}\,dx = \frac{x^3}{3} + \frac{x^2}{2} + 2x + 2\ln|x - 1| + C.

Check the division: $(x - 1)(x^{2} + x + 2) = x^{3} + x^{2} + 2x - x^{2} - x - 2 = x^{3} + x - 2$ , and adding the remainder $2$ gives $x^{3} + x$ .

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Example 5.38 (A repeated linear factor). Compute $\int\frac{3x + 2}{x(x + 1)^2}\,dx$ .

Solution.

The fraction is proper. The factor $(x + 1)$ is repeated, so the decomposition has three terms:

\frac{3x + 2}{x(x + 1)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}.

Cover-up gives the two coefficients it can reach. Covering $x$ and setting $x = 0$ : $A = \frac{2}{1} = 2$ . Covering $(x + 1)^2$ and setting $x = -1$ : $C = \frac{-3 + 2}{-1} = 1$ .
For $B$ , clear denominators: $3x + 2 = A(x + 1)^2 + Bx(x + 1) + Cx$ . The coefficient of $x^{2}$ on the right is $A + B$ , and on the left it is $0$ , so $B = -A = -2$ .
Verify at $x = 1$ : the left side is $5$ ; the right side is $2\cdot 4 - 2\cdot 2 + 1 = 5$ .
Integrate:

\int\frac{3x + 2}{x(x + 1)^2}\,dx = 2\ln|x| - 2\ln|x + 1| - \frac{1}{x + 1} + C.

The last term comes from $\int\frac{dx}{(x + 1)^2} = -\frac{1}{x + 1}$ ; it is an algebraic term, not a logarithm, and forgetting it is the characteristic error of the repeated case.

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Example 5.39 (An irreducible quadratic factor). Compute $\int\frac{2x^{2} - x + 4}{x^{3} + 4x}\,dx$ .

Solution.

Factor: $x^{3} + 4x = x(x^{2} + 4)$ , and $x^{2} + 4$ does not factor over the reals.
The decomposition is

\frac{2x^2 - x + 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}.

Cover-up for $A$ : set $x = 0$ with $x$ covered, $A = \frac{4}{4} = 1$ .
Clear denominators: $2x^{2} - x + 4 = A(x^{2} + 4) + (Bx + C)x = (A + B)x^{2} + Cx + 4A$ . Comparing coefficients: $A + B = 2$ gives $B = 1$ , and $C = -1$ .
Integrate. Split the quadratic piece into a logarithm part and an arctangent part:

\int\Big(\frac1x + \frac{x}{x^2 + 4} - \frac{1}{x^2 + 4}\Big)dx = \ln|x| + \frac12\ln(x^2 + 4) - \frac12\arctan\frac{x}{2} + C.

Check at $x = 2$ : the original fraction is $\frac{8 - 2 + 4}{8 + 8} = \frac{10}{16} = \frac58$ ; the decomposition gives $\frac12 + \frac{2 - 1}{8} = \frac58$ .

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Example 5.40 (A repeated irreducible quadratic: setting it up). Write out the decomposition of $\dfrac{1 - x + 2x^{2} - x^{3}}{x(x^{2} + 1)^2}$ and determine its coefficients.

Solution.

The denominator has a simple linear factor $x$ and a repeated irreducible quadratic $(x^{2} + 1)^2$ , so the form is

\frac{1 - x + 2x^2 - x^3}{x(x^2 + 1)^2} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} + \frac{Dx + E}{(x^2 + 1)^2}.

Cover-up for $A$ : $A = \frac{1}{1} = 1$ .
Clear denominators and expand:

1 - x + 2x^2 - x^3 = A(x^2 + 1)^2 + (Bx + C)x(x^2 + 1) + (Dx + E)x = (A + B)x^4 + Cx^3 + (2A + B + D)x^2 + (C + E)x + A.

Compare coefficients. $x^{4}$ : $A + B = 0$ , so $B = -1$ . $x^{3}$ : $C = -1$ . $x^{2}$ : $2A + B + D = 2$ , so $D = 2 - 2 + 1 = 1$ . $x$ : $C + E = -1$ , so $E = 0$ . Constant: $A = 1$ , consistent.
The decomposition is

\frac{1}{x} - \frac{x + 1}{x^2 + 1} + \frac{x}{(x^2 + 1)^2}.

Each piece is now standard: the first is $\ln|x|$ , the second splits into $-\frac12\ln(x^{2} + 1) - \arctan x$ , and the third is $-\frac{1}{2(x^{2} + 1)}$ by the substitution $u = x^{2} + 1$ . Had the last numerator contained a constant term, the integral $\int\frac{dx}{(x^{2} + 1)^2}$ would have required the substitution $x = \tan\theta$ and reduced to $\int\cos^{2}\theta\,d\theta$ .

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Pitfall. Three set-up errors account for most wrong answers. Not dividing first when the fraction is improper: the decomposition theorem is false for improper fractions, and the unknown coefficients will have no solution. Writing only $\frac{A}{(x - r)^2}$ for a repeated factor: the theorem requires a term for every power from $1$ up to $k$ . Writing $\frac{A}{x^{2} + 4}$ instead of $\frac{Ax + B}{x^{2} + 4}$ : the numerator over a quadratic is linear, and dropping the $Ax$ loses the logarithmic part of the answer.

Exercises for §5.5 →

5.6Rationalising substitutions

Partial fractions handle rational functions, and a large class of integrands can be made rational by a substitution that removes a root. If the integrand contains $\sqrt[n]{g(x)}$ for a simple $g$ , the substitution $u = \sqrt[n]{g(x)}$ replaces the root by the plain variable $u$ , and when $g$ is linear, $x$ and $dx$ become polynomials in $u$ .

Method 5.41 (Rationalising substitution).

If the integrand contains a single root $\sqrt[n]{ax + b}$ , put $u = \sqrt[n]{ax + b}$ , so $x = \frac{u^n - b}{a}$ and $dx = \frac{n u^{n-1}}{a}\,du$ .
If it contains several roots of the same expression, $\sqrt[m]{x}$ and $\sqrt[n]{x}$ say, put $u = x^{1/k}$ where $k$ is the least common multiple of $m$ and $n$ ; every root is then an integer power of $u$ .
The result is a rational function of $u$ . Integrate it by partial fractions (after long division if it is improper) and substitute back.

Intuition. The obstacle in $\int\frac{\sqrt{x + 4}}{x}\,dx$ is the root, not the $x$ . Rename the root: call it $u$ . Then $x$ is $u^{2} - 4$ , which is a polynomial, and $dx$ is $2u\,du$ , also a polynomial, and suddenly every part of the integrand is built from $u$ with the four arithmetic operations. Partial fractions can take it from there. The substitution does not make the integral smaller; it makes it rational, which is the one kind of integral this chapter can always finish.

Example 5.42 (A square root of a linear expression). Compute $\int\frac{\sqrt{x + 4}}{x}\,dx$ .

Solution.

Put $u = \sqrt{x + 4}$ , so $x = u^{2} - 4$ and $dx = 2u\,du$ .
Substitute:

\int\frac{u}{u^2 - 4}\cdot 2u\,du = 2\int\frac{u^2}{u^2 - 4}\,du.

The fraction is improper. Dividing, $\frac{u^{2}}{u^{2} - 4} = 1 + \frac{4}{u^{2} - 4}$ , and by partial fractions $\frac{4}{u^{2} - 4} = \frac{1}{u - 2} - \frac{1}{u + 2}$ (cover-up: $\frac{4}{2 + 2} = 1$ and $\frac{4}{-2 - 2} = -1$ ).
Integrate:

2\int\Big(1 + \frac{1}{u - 2} - \frac{1}{u + 2}\Big)du = 2u + 2\ln\Big|\frac{u - 2}{u + 2}\Big| + C.

Back in $x$ :

\int\frac{\sqrt{x + 4}}{x}\,dx = 2\sqrt{x + 4} + 2\ln\left|\frac{\sqrt{x + 4} - 2}{\sqrt{x + 4} + 2}\right| + C.

Sanity check: for large $x$ the integrand is about $x^{-1/2}$ , whose antiderivative $2\sqrt{x}$ matches the leading term $2\sqrt{x + 4}$ ; the logarithm tends to $\ln 1 = 0$ .

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Example 5.43 (A root in the denominator). Compute $\int\frac{dx}{1 + \sqrt{x}}$ .

Solution.

Put $u = \sqrt{x}$ , so $x = u^{2}$ and $dx = 2u\,du$ .
The integral becomes $\int\frac{2u}{1 + u}\,du$ , an improper rational function. Divide: $\frac{2u}{1 + u} = 2 - \frac{2}{1 + u}$ .
Integrate: $2u - 2\ln|1 + u| + C$ .
Back in $x$ : $\int\frac{dx}{1 + \sqrt{x}} = 2\sqrt{x} - 2\ln(1 + \sqrt{x}) + C$ . The absolute value is unnecessary since $1 + \sqrt{x} > 0$ .
Check: the derivative is $\frac{1}{\sqrt{x}} - \frac{2}{1 + \sqrt{x}}\cdot\frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}\Big(1 - \frac{1}{1 + \sqrt{x}}\Big) = \frac{1}{\sqrt{x}}\cdot\frac{\sqrt{x}}{1 + \sqrt{x}} = \frac{1}{1 + \sqrt{x}}$ .

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Example 5.44 (Two different roots). Compute $\int\frac{dx}{\sqrt{x} + \sqrt[3]{x}}$ .

Solution.

The roots have indices $2$ and $3$ , with least common multiple $6$ . Put $u = x^{1/6}$ , so $x = u^{6}$ , $dx = 6u^{5}\,du$ , $\sqrt{x} = u^{3}$ and $\sqrt[3]{x} = u^{2}$ .
Substitute:

\int\frac{6u^5}{u^3 + u^2}\,du = 6\int\frac{u^3}{u + 1}\,du.

Divide: $u^{3} = (u + 1)(u^{2} - u + 1) - 1$ , so $\frac{u^{3}}{u + 1} = u^{2} - u + 1 - \frac{1}{u + 1}$ .
Integrate:

6\Big(\frac{u^3}{3} - \frac{u^2}{2} + u - \ln|u + 1|\Big) + C = 2u^3 - 3u^2 + 6u - 6\ln(u + 1) + C.

Back in $x$ :

\int\frac{dx}{\sqrt{x} + \sqrt[3]{x}} = 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln\big(\sqrt[6]{x} + 1\big) + C.

Sanity check: for large $x$ the integrand is about $x^{-1/2}$ and the answer's leading term is $2\sqrt{x}$ , as it should be.

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Remark (The substitution $t = \tan(x/2)$ ). A rational function of $\sin x$ and $\cos x$ can also be rationalised. With $t = \tan\frac{x}{2}$ one has

\sin x = \frac{2t}{1 + t^2}, \qquad \cos x = \frac{1 - t^2}{1 + t^2}, \qquad dx = \frac{2\,dt}{1 + t^2},

so any integrand built rationally from $\sin x$ and $\cos x$ becomes a rational function of $t$ . For instance $\int\frac{dx}{1 + \sin x}$ becomes $\int\frac{2\,dt}{(1 + t)^2} = -\frac{2}{1 + t} + C = -\frac{2}{1 + \tan(x/2)} + C$ . The substitution always works, and is usually the last resort because the resulting rational function can be unpleasant.

Pitfall. After substituting $u = \sqrt{x + 4}$ , every $x$ must go, including the one hiding in $dx$ . Writing $\int\frac{u}{x}\,dx$ with $x$ and $dx$ still present is the standard half-finished substitution, and it cannot be integrated. Express $x$ and $dx$ in terms of $u$ before doing anything else.

Exercises for §5.6 →

5.7Strategy for integration

Each technique so far came with its own recognisable shape. In practice the shape is not announced, and the skill being tested is deciding which technique to try. That decision starts from the table of standard forms, since every technique is a way of transforming an integral into one of them.

\int x^n\,dx = \frac{x^{n+1}}{n+1}\ (n \ne -1), \qquad \int\frac{dx}{x} = \ln|x|, \qquad \int e^x\,dx = e^x, \qquad \int a^x\,dx = \frac{a^x}{\ln a},

\int\sin x\,dx = -\cos x, \qquad \int\cos x\,dx = \sin x, \qquad \int\sec^2 x\,dx = \tan x, \qquad \int\csc^2 x\,dx = -\cot x,

\int\sec x\tan x\,dx = \sec x, \qquad \int\csc x\cot x\,dx = -\csc x, \qquad \int\tan x\,dx = \ln|\sec x|, \qquad \int\cot x\,dx = \ln|\sin x|,

\int\sec x\,dx = \ln|\sec x + \tan x|, \qquad \int\csc x\,dx = \ln|\csc x - \cot x|, \qquad \int\sinh x\,dx = \cosh x, \qquad \int\cosh x\,dx = \sinh x,

\int\frac{dx}{x^2 + a^2} = \frac1a\arctan\frac{x}{a}, \qquad \int\frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a}, \qquad \int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x - a}{x + a}\right|, \qquad \int\frac{dx}{\sqrt{x^2 \pm a^2}} = \ln\left|x + \sqrt{x^2 \pm a^2}\right|.

Each formula carries an implicit $+C$ . The last row is where trigonometric substitution and partial fractions deliver their results, and knowing those four outright saves redoing the substitution every time.

Method 5.45 (The four-step strategy).

Simplify the integrand. Multiply out, cancel, use trigonometric identities ( $\tan^{2} = \sec^{2} - 1$ , products into sums, $\sin x\cos x = \tfrac12\sin 2x$ ), rationalise, or split a fraction whose numerator is a sum. Many integrals are only hard as written.
Look for an obvious substitution. Is there a function $g(x)$ whose derivative $g'(x)$ appears as a factor, up to a constant? If so, $u = g(x)$ probably finishes the job.
Classify by form. A rational function calls for partial fractions after long division. A product of a polynomial with $e^x$ , $\sin x$ , $\cos x$ , $\ln x$ or an inverse trigonometric function calls for parts, as does a lone $\ln x$ or $\arctan x$ . Powers of trigonometric functions call for the parity rules. A root of a quadratic calls for trigonometric substitution after completing the square; a root of a linear expression calls for a rationalising substitution.
Try again. Substitute and then integrate by parts; manipulate the integrand by multiplying by a conjugate or by $1$ in disguise; relate the integral to one already done; or combine two methods. Most integrals that resist the first three steps yield to a chain of two techniques.

Method 5.46 (Which technique first).

Is it in the table, or in the table after one algebraic rewrite? Done.
Is there a factor that is the derivative of something inside? Substitute.
Is it $\frac{P(x)}{Q(x)}$ ? Divide if improper, then partial fractions.
Is it a product of two functions from different families, or a lone logarithm or inverse trigonometric function? Parts.
Is it $\sin^m x\cos^n x$ or $\tan^m x\sec^n x$ , or a product of sines and cosines of different frequencies? Parity rules or product-to-sum.
Does it contain $\sqrt{a^{2} - x^{2}}$ , $\sqrt{a^{2} + x^{2}}$ or $\sqrt{x^{2} - a^{2}}$ , possibly after completing the square? Trigonometric substitution.
Does it contain $\sqrt[n]{ax + b}$ , or is it rational in $\sin x$ and $\cos x$ ? Rationalise with $u = \sqrt[n]{ax + b}$ or $t = \tan(x/2)$ .
None of the above? Rewrite the integrand and start from step 1, or suspect that no elementary antiderivative exists.

Intuition. Faced with $\int\frac{\tan^{3} x}{\cos^{3} x}\,dx$ , a student who has memorised the six techniques as six separate boxes sees nothing: it is not a product of a polynomial and an exponential, there is no square root, it is not rational. The strategy says to start by rewriting. Since $\frac{1}{\cos^{3} x} = \sec^{3} x$ , the integrand is $\tan^{3} x\sec^{3} x$ , and now step 5 lights up: odd power of tangent, save $\sec x\tan x$ . The technique was always the right one; the integrand simply had not been written in the form that reveals it.

Example 5.47 (Rewrite, then classify). Compute $\int\frac{\tan^{3} x}{\cos^{3} x}\,dx$ .

Solution.

Step 1, simplify: $\frac{\tan^{3} x}{\cos^{3} x} = \tan^{3} x\sec^{3} x$ .
Step 3, classify: an odd power of tangent with a power of secant. Save $\sec x\tan x$ and convert the remaining $\tan^{2} x = \sec^{2} x - 1$ :

\int\tan^3 x\sec^3 x\,dx = \int(\sec^2 x - 1)\sec^2 x\cdot\sec x\tan x\,dx.

With $u = \sec x$ , $du = \sec x\tan x\,dx$ , this is $\int(u^{4} - u^{2})\,du = \frac{u^{5}}{5} - \frac{u^{3}}{3} + C$ .
So $\int\frac{\tan^{3} x}{\cos^{3} x}\,dx = \frac{\sec^{5} x}{5} - \frac{\sec^{3} x}{3} + C$ .
Check: the derivative is $\sec^{4} x\cdot\sec x\tan x - \sec^{2} x\cdot\sec x\tan x = \sec^{3} x\tan x(\sec^{2} x - 1) = \sec^{3} x\tan^{3} x$ .

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Example 5.48 (Substitution followed by parts). Compute $\int e^{\sqrt{x}}\,dx$ .

Solution.

No step-1 rewrite helps and no factor is the derivative of $\sqrt{x}$ . Step 4: try a substitution anyway to remove the root, $u = \sqrt{x}$ , $x = u^{2}$ , $dx = 2u\,du$ :

\int e^{\sqrt{x}}\,dx = \int 2u e^u\,du.

This is a polynomial times an exponential: parts with $u$ as the polynomial factor (rename to avoid a clash: differentiate $2u$ , integrate $e^u$ ):

\int 2u e^u\,du = 2u e^u - \int 2e^u\,du = 2u e^u - 2e^u + C.

Back in $x$ : $\int e^{\sqrt{x}}\,dx = 2(\sqrt{x} - 1)e^{\sqrt{x}} + C$ .
Check: differentiating $2(\sqrt{x} - 1)e^{\sqrt{x}}$ by the product and chain rules gives $\frac{1}{\sqrt{x}}e^{\sqrt{x}} + 2(\sqrt{x} - 1)e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}} = e^{\sqrt{x}}\Big(\frac{1}{\sqrt x} + 1 - \frac{1}{\sqrt x}\Big) = e^{\sqrt{x}}$ .

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Example 5.49 (Rationalise the integrand). Compute $\int\sqrt{\frac{1 - x}{1 + x}}\,dx$ .

Solution.

The root of a quotient of linear expressions fits no pattern directly. Step 4: multiply inside the root by $\frac{1 - x}{1 - x}$ to make the denominator a difference of squares:

\sqrt{\frac{1 - x}{1 + x}} = \sqrt{\frac{(1 - x)^2}{1 - x^2}} = \frac{1 - x}{\sqrt{1 - x^2}} \quad (-1 < x \le 1).

Split the fraction: $\frac{1}{\sqrt{1 - x^{2}}} - \frac{x}{\sqrt{1 - x^{2}}}$ .
The first piece is in the table; the second is a substitution with $w = 1 - x^{2}$ :

\int\sqrt{\frac{1 - x}{1 + x}}\,dx = \arcsin x + \sqrt{1 - x^2} + C.

Check: the derivative is $\frac{1}{\sqrt{1 - x^{2}}} - \frac{x}{\sqrt{1 - x^{2}}} = \frac{1 - x}{\sqrt{1 - x^{2}}}$ , which is the rewritten integrand.

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Example 5.50 (Multiply by a conjugate). Compute $\int\frac{dx}{1 + \cos x}$ .

Solution.

A rational function of $\cos x$ ; the substitution $t = \tan(x/2)$ would work but is heavy. Step 4: multiply numerator and denominator by $1 - \cos x$ :

\frac{1}{1 + \cos x} = \frac{1 - \cos x}{1 - \cos^2 x} = \frac{1 - \cos x}{\sin^2 x} = \csc^2 x - \csc x\cot x.

Both terms are in the table:

\int\frac{dx}{1 + \cos x} = -\cot x + \csc x + C.

Check: the derivative of $-\cot x + \csc x$ is $\csc^{2} x - \csc x\cot x = \frac{1 - \cos x}{\sin^{2} x} = \frac{1}{1 + \cos x}$ .
The half-angle identity $1 + \cos x = 2\cos^{2}(x/2)$ gives the same integral as $\tan(x/2) + C$ ; the two answers agree because $\csc x - \cot x = \frac{1 - \cos x}{\sin x} = \tan\frac{x}{2}$ .

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Remark (Can every function be integrated?). The Fundamental Theorem guarantees that every continuous $f$ has an antiderivative, namely $F(x) = \int_a^x f(t)\,dt$ . The question is whether $F$ can be written as an elementary function, one built from polynomials, roots, exponentials, logarithms, trigonometric functions and their inverses by arithmetic and composition. For

e^{x^2}, \qquad e^{-x^2}, \qquad \frac{e^x}{x}, \qquad \sin(x^2), \qquad \frac{\sin x}{x}, \qquad \frac{1}{\ln x}, \qquad \sqrt{x^3 + 1}

the answer is no: Liouville proved in the 1830s that these antiderivatives are not elementary, and no technique, however clever, will produce a formula. Their antiderivatives are nonetheless perfectly good functions with their own names and tables; $\frac{2}{\sqrt\pi}\int_0^x e^{-t^{2}}\,dt$ is the error function of statistics. When a method fails on such an integrand, the failure is not yours. Notice that a single extra factor can change everything: $\int xe^{-x^{2}}\,dx = -\tfrac12 e^{-x^{2}} + C$ is elementary because the $x$ is the derivative of the exponent.

Remark (Tables and computer algebra). Tables of integrals list a few hundred forms with parameters, such as $\int\frac{du}{u^{2}\sqrt{a^{2} + u^{2}}} = -\frac{\sqrt{a^{2} + u^{2}}}{a^{2}u} + C$ . Using one is itself a substitution problem: you must bring your integral into the table's shape, with the right $u$ and $a$ , and remember the factor from $du$ . Computer algebra systems do the same thing at scale, and their answers deserve two checks. First, the form may differ from yours by a constant or by an identity: a system may return $-\tfrac12\cos^{2} x$ where you found $\tfrac12\sin^{2} x$ , or $\operatorname{arctanh}$ where you found a logarithm. Second, systems routinely omit $+C$ and the absolute value in $\ln|x|$ , and may return an expression valid only on part of the domain. Differentiating the output is the universal check, and the strategy in this section is what lets you read the output critically.

Exercises for §5.7 →

5.8Improper integrals

The definite integral $\int_a^b f(x)\,dx$ was defined for a bounded function on a bounded interval. Two natural extensions relax those conditions: the interval may reach to infinity, or the function may blow up at a point of the interval. In both cases the integral is defined as a limit of ordinary integrals, and the new question is whether the limit exists.

Definition 5.51 (Improper integrals of Type 1: infinite intervals).

If $\int_a^t f(x)\,dx$ exists for every $t \ge a$ , then

\int_a^{\infty} f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx,

provided the limit exists as a finite number.

If $\int_t^b f(x)\,dx$ exists for every $t \le b$ , then $\displaystyle\int_{-\infty}^b f(x)\,dx = \lim_{t\to-\infty}\int_t^b f(x)\,dx$ , provided the limit exists.

An improper integral is called convergent if the limit exists and divergent otherwise.

If both $\int_a^{\infty} f$ and $\int_{-\infty}^a f$ converge, then

\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^a f(x)\,dx + \int_a^{\infty} f(x)\,dx,

and the value does not depend on the choice of $a$ .

Definition 5.52 (Improper integrals of Type 2: unbounded integrands).

If $f$ is continuous on $[a, b)$ and unbounded near $b$ , then

\int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx,

provided the limit exists as a finite number.

If $f$ is continuous on $(a, b]$ and unbounded near $a$ , then $\displaystyle\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx$ , provided the limit exists.
If $f$ is unbounded near an interior point $c$ of $(a, b)$ and both $\int_a^c f$ and $\int_c^b f$ converge, then $\int_a^b f = \int_a^c f + \int_c^b f$ . If either piece diverges, so does $\int_a^b f$ .

The definitions say the same thing twice: stop short of the trouble, integrate, and let the stopping point approach the trouble. For Type 1 the trouble is at infinity, for Type 2 at a finite point where $f$ has a vertical asymptote. The third clause in each definition is the important one: an integral with two problems, or a problem in the middle, must be split into pieces each of which has a single problem at one end, and every piece must converge separately. A limit taken symmetrically, such as $\lim_{t\to\infty}\int_{-t}^t f$ , is not the definition and can give a finite answer where the improper integral diverges.

Theorem 5.53 (The $p$ -integral at infinity).

\int_1^{\infty}\frac{dx}{x^p} \text{ converges if } p > 1 \text{ and diverges if } p \le 1. \text{ When it converges its value is } \frac{1}{p - 1}.

Proof. For $p \ne 1$ ,

\int_1^t\frac{dx}{x^p} = \Big[\frac{x^{1-p}}{1 - p}\Big]_1^t = \frac{t^{1-p} - 1}{1 - p}.

If $p > 1$ then $1 - p < 0$ , so $t^{1-p} \to 0$ as $t \to \infty$ and the limit is $\frac{-1}{1 - p} = \frac{1}{p - 1}$ . If $p < 1$ then $1 - p > 0$ , so $t^{1-p} \to \infty$ and the integral diverges. For $p = 1$ , $\int_1^t\frac{dx}{x} = \ln t \to \infty$ , so it diverges as well.∎

Theorem 5.54 (The $p$ -integral at zero).

\int_0^1\frac{dx}{x^p} \text{ converges if } p < 1 \text{ and diverges if } p \ge 1. \text{ When it converges its value is } \frac{1}{1 - p}.

Proof. For $p \ne 1$ ,

\int_t^1\frac{dx}{x^p} = \Big[\frac{x^{1-p}}{1 - p}\Big]_t^1 = \frac{1 - t^{1-p}}{1 - p}.

If $p < 1$ then $t^{1-p} \to 0$ as $t \to 0^+$ and the limit is $\frac{1}{1 - p}$ . If $p > 1$ then $t^{1-p} = 1/t^{p-1} \to \infty$ and the integral diverges. For $p = 1$ , $\int_t^{1}\frac{dx}{x} = -\ln t \to \infty$ .∎

The two theorems are mirror images. Near infinity the integrand must decay faster than $1/x$ ; near zero it must blow up more slowly than $1/x$ . The borderline $1/x$ fails at both ends, because $\ln$ is unbounded in both directions. These are the benchmarks against which every other improper integral is measured in the next section.

Intuition. Paint an infinitely long strip of width $f(x)$ starting at $x = 1$ . With $f(x) = 1/x^{2}$ , painting up to $x = t$ uses $1 - 1/t$ litres: up to $x = 10^6$ that is $0.999999$ litres, and no matter how far you go you never need a full litre. With $f(x) = 1/x$ the same job uses $\ln t$ litres: $13.8$ litres by $x = 10^6$ , $27.6$ by $x = 10^{12}$ , and the number keeps growing, slowly but without bound. Both strips get thinner and thinner. Only one of them has a finite total area.

Example 5.55 (A limit that needs L'Hôpital). Evaluate $\int_{-\infty}^0 xe^x\,dx$ .

Solution.

This is Type 1 with the infinite end on the left: compute $\int_t^{0} xe^x\,dx$ and let $t \to -\infty$ .
By parts, $\int xe^x\,dx = xe^x - e^x = (x - 1)e^x$ , so

\int_t^0 xe^x\,dx = \Big[(x - 1)e^x\Big]_t^0 = -1 - (t - 1)e^t.

As $t \to -\infty$ , $e^t \to 0$ while $t - 1 \to -\infty$ , an indeterminate product. Write it as $\frac{t - 1}{e^{-t}}$ and apply L'Hôpital's Rule: $\lim_{t\to-\infty}\frac{1}{-e^{-t}} = 0$ .
So $\int_{-\infty}^0 xe^x\,dx = -1$ .
Sanity check: the integrand is negative on $(-\infty, 0)$ and the exponential crushes the linear factor, so a finite negative answer is expected.

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Example 5.56 (Both ends infinite). Evaluate $\int_{-\infty}^{\infty}\frac{dx}{1 + x^{2}}$ .

Solution.

Split at $0$ : the integral is $\int_{-\infty}^0\frac{dx}{1 + x^{2}} + \int_0^{\infty}\frac{dx}{1 + x^{2}}$ , and both must converge.
$\int_0^t\frac{dx}{1 + x^{2}} = \arctan t \to \frac{\pi}{2}$ as $t \to \infty$ .
$\int_t^{0}\frac{dx}{1 + x^{2}} = -\arctan t \to \frac{\pi}{2}$ as $t \to -\infty$ .
Both converge, so $\int_{-\infty}^{\infty}\frac{dx}{1 + x^{2}} = \frac{\pi}{2} + \frac{\pi}{2} = \pi$ .
Sanity check: the integrand is at most $1$ and decays like $1/x^{2}$ , so a finite total is expected, and $\pi \approx 3.14$ is a plausible area under a bump of height $1$ and width a few units.

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Example 5.57 (Type 2 at an endpoint). Evaluate $\int_2^5\frac{dx}{\sqrt{x - 2}}$ .

Solution.

The integrand is unbounded as $x \to 2^+$ , so this is Type 2 with the trouble at the left endpoint: compute $\int_t^{5}(x - 2)^{-1/2}\,dx$ and let $t \to 2^+$ .
$\int_t^{5}(x - 2)^{-1/2}\,dx = \Big[2\sqrt{x - 2}\Big]_t^{5} = 2\sqrt3 - 2\sqrt{t - 2}$ .
As $t \to 2^+$ the second term tends to $0$ , so the integral converges and equals $2\sqrt3$ .
Sanity check: with $u = x - 2$ this is $\int_0^3 u^{-1/2}\,du$ , a $p$ -integral at zero with $p = \tfrac12 < 1$ , which converges.

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Example 5.58 (Type 2 that diverges). Determine whether $\int_0^{\pi/2}\sec x\,dx$ converges.

Solution.

$\sec x \to \infty$ as $x \to (\pi/2)^-$ , so compute $\int_0^t\sec x\,dx$ and let $t \to (\pi/2)^-$ .
$\int_0^t\sec x\,dx = \Big[\ln|\sec x + \tan x|\Big]_0^t = \ln(\sec t + \tan t) - \ln 1$ .
As $t \to (\pi/2)^-$ , both $\sec t$ and $\tan t$ tend to $+\infty$ , so the logarithm tends to $\infty$ .
The integral diverges.
Sanity check: near $\pi/2$ , $\sec x \approx \frac{1}{\pi/2 - x}$ , a $p = 1$ singularity, which is the divergent borderline.

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Example 5.59 (The trap: a singularity in the middle). Evaluate $\int_0^3\frac{dx}{x - 1}$ if it converges.

Solution.

A student who does not look at the integrand writes $\Big[\ln|x - 1|\Big]_0^3 = \ln 2 - \ln 1 = \ln 2$ . This is wrong: the integrand has a vertical asymptote at $x = 1$ , inside the interval, and the Fundamental Theorem does not apply across a discontinuity.
By the definition, split at $x = 1$ and test each piece. The left piece is

\int_0^1\frac{dx}{x - 1} = \lim_{t\to 1^-}\Big[\ln|x - 1|\Big]_0^t = \lim_{t\to 1^-}\ln(1 - t) - \ln 1 = -\infty.

One piece diverges, so $\int_0^3\frac{dx}{x - 1}$ diverges. There is no need to examine the right piece.
The value $\ln 2$ obtained by ignoring the singularity is meaningless: the same careless method gives $\int_{-1}^1\frac{dx}{x^{2}} = \Big[-\frac1x\Big]_{-1}^1 = -2$ , a negative area for a positive function.

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Example 5.60 (A logarithmic singularity). Evaluate $\int_0^1\ln x\,dx$ .

Solution.

$\ln x \to -\infty$ as $x \to 0^+$ , so compute $\int_t^{1}\ln x\,dx$ and let $t \to 0^+$ .
By parts, $\int\ln x\,dx = x\ln x - x$ , so

\int_t^1\ln x\,dx = \Big[x\ln x - x\Big]_t^1 = -1 - t\ln t + t.

$t \to 0$ directly, and $t\ln t \to 0$ by L'Hôpital's Rule applied to $\frac{\ln t}{1/t}$ , as in the Limits chapter.
So $\int_0^1\ln x\,dx = -1$ .
Sanity check: the region lies below the axis, and its area is finite because $\ln x$ blows up only logarithmically, far more slowly than any power $1/x^p$ .

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Pitfall. Always look at the integrand on the whole interval before applying the Fundamental Theorem. A denominator that vanishes inside $[a, b]$ , a $\ln$ or $\tan$ or $\sec$ that reaches its asymptote, a negative power of $x$ on an interval containing $0$ : each makes the integral improper, and each must be split at the bad point. And never replace $\int_{-\infty}^{\infty}$ by $\lim_{t\to\infty}\int_{-t}^t$ : the latter is $0$ for $f(x) = x$ by symmetry, but $\int_0^{\infty}x\,dx$ diverges, so $\int_{-\infty}^{\infty}x\,dx$ diverges.

Exercises for §5.8 →

5.9Comparison theorem for improper integrals

Often it is impossible to find an antiderivative, so the limit in the definition cannot be computed, yet the question of convergence can still be settled by comparing with an integral that is understood. For non-negative integrands the principle is simple: a smaller function cannot enclose more area than a larger one.

Theorem 5.61 (Comparison theorem). Let $f$ and $g$ be continuous with $0 \le g(x) \le f(x)$ for all $x \ge a$ .

If $\int_a^{\infty} f(x)\,dx$ converges, then $\int_a^{\infty} g(x)\,dx$ converges, and $\int_a^{\infty} g \le \int_a^{\infty} f$ .
If $\int_a^{\infty} g(x)\,dx$ diverges, then $\int_a^{\infty} f(x)\,dx$ diverges.

The same statements hold for Type 2 integrals on $[a, b)$ or $(a, b]$ with the inequality assumed on that interval.

Proof. Let $G(t) = \int_a^t g(x)\,dx$ . Because $g \ge 0$ , $G$ is an increasing function of $t$ . Because $g \le f$ , $G(t) \le \int_a^t f(x)\,dx \le \int_a^{\infty} f(x)\,dx$ , so $G$ is bounded above by a fixed number. An increasing function that is bounded above has a finite limit as $t \to \infty$ ; this is the completeness property of the real numbers, whose proof belongs to real analysis. That limit is $\int_a^{\infty} g$ , and the inequality is inherited from $G(t) \le \int_a^{\infty} f$ .

Part 2 is the contrapositive of part 1: if $\int_a^{\infty} f$ converged, part 1 would force $\int_a^{\infty} g$ to converge too.∎

The hypothesis $g \ge 0$ cannot be dropped. Without it, $G$ need not be increasing and a bounded $G$ could oscillate forever. The direction of the conclusions is the thing to get right: a convergent big integral squeezes a small one into convergence; a divergent small integral pushes a big one into divergence. The other two combinations tell you nothing: if the small integral converges, the big one may do either, and if the big one diverges, the small one may do either.

Corollary 5.62 (Absolute convergence). If $f$ is continuous on $[a, \infty)$ and $\int_a^{\infty}|f(x)|\,dx$ converges, then $\int_a^{\infty} f(x)\,dx$ converges.

Proof. The function $f + |f|$ satisfies $0 \le f + |f| \le 2|f|$ , so $\int_a^{\infty}(f + |f|)$ converges by comparison with $2|f|$ . Then $\int_a^{\infty} f = \int_a^{\infty}(f + |f|) - \int_a^{\infty}|f|$ is a difference of two convergent integrals.∎

Proposition 5.63 (Limit comparison). Let $f, g$ be continuous and positive for $x \ge a$ , with $\lim_{x\to\infty}\frac{f(x)}{g(x)} = L$ where $0 < L < \infty$ . Then $\int_a^{\infty} f$ and $\int_a^{\infty} g$ either both converge or both diverge.

Proof. For all large $x$ , say $x \ge b$ , the ratio lies between $\tfrac{L}{2}$ and $2L$ , so $\tfrac{L}{2}g(x) \le f(x) \le 2L\,g(x)$ . Comparison on $[b, \infty)$ in each direction gives the claim, and the integrals over $[a, b]$ are finite and do not affect convergence.∎

Intuition. You want to know whether $\int_1^{\infty}e^{-x^{2}}\,dx$ is finite, and you cannot integrate $e^{-x^{2}}$ . But for $x \ge 1$ you have $x^{2} \ge x$ , so $e^{-x^{2}} \le e^{-x}$ , and $\int_1^{\infty}e^{-x}\,dx = e^{-1}$ is an easy number. The graph of $e^{-x^{2}}$ lies entirely under a curve whose total area is $1/e \approx 0.37$ , so the area under $e^{-x^{2}}$ is finite and at most $0.37$ . You never learn its exact value from this argument, but you learn the only thing the question asked.

Example 5.64 (Convergence by comparison). Show that $\int_1^{\infty}e^{-x^{2}}\,dx$ converges.

Solution.

There is no elementary antiderivative, so the definition cannot be applied directly.
For $x \ge 1$ , $x^{2} \ge x$ , hence $-x^{2} \le -x$ and $0 \le e^{-x^{2}} \le e^{-x}$ .
The larger integral converges: $\int_1^{\infty}e^{-x}\,dx = \lim_{t\to\infty}\big(e^{-1} - e^{-t}\big) = e^{-1}$ .
By part 1 of the comparison theorem, $\int_1^{\infty}e^{-x^{2}}\,dx$ converges, with value at most $e^{-1}$ .
Consequently $\int_0^{\infty}e^{-x^{2}}\,dx = \int_0^1 e^{-x^{2}}\,dx + \int_1^{\infty}e^{-x^{2}}\,dx$ also converges; the first piece is an ordinary integral. Its exact value, $\frac{\sqrt\pi}{2}$ , is found by a double integral in the chapter on multiple integrals.

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Example 5.65 (Divergence by comparison). Show that $\int_1^{\infty}\frac{1 + e^{-x}}{x}\,dx$ diverges.

Solution.

Since $e^{-x} > 0$ , the integrand satisfies $\frac{1 + e^{-x}}{x} \ge \frac{1}{x} > 0$ for $x \ge 1$ .
The smaller integral $\int_1^{\infty}\frac{dx}{x}$ diverges, being the $p$ -integral with $p = 1$ .
By part 2 of the comparison theorem, $\int_1^{\infty}\frac{1 + e^{-x}}{x}\,dx$ diverges.
Sanity check: the integrand tends to $1/x$ at infinity, which is exactly the borderline that fails; the extra $e^{-x}$ only adds area.

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Example 5.66 (Absolute convergence with an oscillating integrand). Show that $\int_1^{\infty}\frac{\sin x}{x^{2}}\,dx$ converges.

Solution.

The integrand changes sign, so the comparison theorem does not apply to it directly.
Its absolute value satisfies $0 \le \frac{|\sin x|}{x^{2}} \le \frac{1}{x^{2}}$ , and $\int_1^{\infty}\frac{dx}{x^{2}} = 1$ converges.
By comparison, $\int_1^{\infty}\frac{|\sin x|}{x^{2}}\,dx$ converges, and by the corollary $\int_1^{\infty}\frac{\sin x}{x^{2}}\,dx$ converges.
Sanity check: the value is bounded in absolute value by $1$ ; numerically it is about $0.50$ .

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Example 5.67 (Comparison for a Type 2 integral). Show that $\int_0^1\frac{e^x}{\sqrt x}\,dx$ converges and find an upper bound for it.

Solution.

The trouble is at $x = 0$ , where $\frac{1}{\sqrt x}$ blows up. On $(0, 1]$ , $1 \le e^x \le e$ , so $0 \le \frac{e^x}{\sqrt x} \le \frac{e}{\sqrt x}$ .
The larger integral is a $p$ -integral at zero with $p = \tfrac12$ : $\int_0^1\frac{e}{\sqrt x}\,dx = e\cdot\frac{1}{1 - 1/2} = 2e$ .
By comparison the given integral converges and $\int_0^1\frac{e^x}{\sqrt x}\,dx \le 2e \approx 5.44$ .
Sanity check: the same comparison from below, $e^x \ge 1$ , gives a lower bound of $\int_0^1\frac{dx}{\sqrt x} = 2$ ; the true value, about $2.93$ , sits between.

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Pitfall. Comparison decides convergence, not value. Writing $\int_1^{\infty}e^{-x^{2}}\,dx = e^{-1}$ because $e^{-x^{2}} \le e^{-x}$ is wrong; the correct statement is $\int_1^{\infty}e^{-x^{2}}\,dx \le e^{-1}$ . And the inequality must point the right way: bounding $\frac{1}{x^{2} + 1}$ below by $0$ proves nothing, since a convergent small integral says nothing about the big one. To prove convergence you need a larger function with a convergent integral; to prove divergence, a smaller function with a divergent integral.

Summary (Which form, which method).

A product of families ( $x^n e^x$ , $x^n\sin x$ , a lone $\ln x$ or $\arctan x$ ): parts, $\int u\,dv=uv-\int v\,du$ , choosing $u$ by LIATE; for a high power, the reduction formulas for $\sin^n x$ , $\cos^n x$ , $\tan^n x$ , $\sec^n x$ , $x^ne^x$ , $(\ln x)^n$ .
$\sin^m x\cos^n x$ : odd $n$ , save $\cos x$ and put $u=\sin x$ ; odd $m$ , save $\sin x$ and put $u=\cos x$ ; both even, half-angle identities. $\tan^m x\sec^n x$ : even $n\ge2$ , save $\sec^{2}x$ , $u=\tan x$ ; odd $m$ , save $\sec x\tan x$ , $u=\sec x$ ; otherwise reduce to $\int\sec x\,dx=\ln|\sec x+\tan x|+C$ and $\int\sec^{3}x\,dx$ . Different frequencies: product-to-sum.
$\sqrt{a^{2}-x^{2}}$ , $\sqrt{a^{2}+x^{2}}$ , $\sqrt{x^{2}-a^{2}}$ (after completing the square): $x=a\sin\theta$ , $a\tan\theta$ , $a\sec\theta$ on the stated ranges, which fix the sign of the root; convert back with a reference triangle.
$P(x)/Q(x)$ : divide if improper, factor $Q$ , decompose by the four cases (repeated and irreducible-quadratic factors included), find coefficients by cover-up, integrate with $\ln$ and $\arctan$ .
$\sqrt[n]{ax+b}$ : $u=\sqrt[n]{ax+b}$ rationalises; several roots, $u=x^{1/k}$ with $k=\operatorname{lcm}$ .
An infinite limit or an unbounded integrand: define by a one-sided limit, split so each piece has a single problem, and compare against $\int_1^{\infty}x^{-p}$ (converges iff $p>1$ ) or $\int_0^1 x^{-p}$ (converges iff $p<1$ ). Comparison needs $0\le g\le f$ .

Exercises for §5.9 →

Common Mistakes

Dropping the minus sign in $\int u\,dv = uv - \int v\,du$ , or choosing $dv$ to be something you cannot integrate, such as $\ln x\,dx$ or $\arctan x\,dx$ .
Switching the roles of $u$ and $dv$ in the second round of a repeated integration by parts. That undoes the first round and returns you to the start.
Writing $\int\sin^{2} x\,dx = \frac{\sin^{3} x}{3}$ . Even powers of sine and cosine need the half-angle identities; there is no $du$ to attach the power rule to.
Applying a reduction formula below its floor: the sine and secant formulas need $n \ge 2$ , and the base cases $\int\sin x$ , $\int\sec x$ , $\int\tan x$ must be known on their own.
Forgetting to replace $dx$ in a trigonometric substitution, or mixing $x$ -limits with a $\theta$ -antiderivative in a definite integral.
Attempting partial fractions on an improper rational function without dividing first. The decomposition does not exist until $\deg P < \deg Q$ .
Omitting the lower powers of a repeated factor ( $\frac{A}{x - r} + \frac{B}{(x - r)^2}$ , not just $\frac{B}{(x - r)^2}$ ), or writing a constant instead of $Ax + B$ over an irreducible quadratic.
Leaving $x$ or $dx$ behind after a rationalising substitution. Every occurrence of $x$ , including the one in $dx$ , must be expressed in terms of $u$ .
Treating an improper integral as an ordinary one. Integrating $\frac{1}{x - 1}$ across $x = 1$ with the Fundamental Theorem gives a finite number that means nothing; the integral diverges.
Evaluating $\int_{-\infty}^{\infty} f$ as $\lim_{t\to\infty}\int_{-t}^t f$ . The definition requires each half to converge on its own.
Concluding from $\int g$ convergent and $g \le f$ that $\int f$ converges, or from $\int f$ divergent that $\int g$ diverges. The comparison theorem runs in one direction only.
Reporting the bound from a comparison as the value of the integral. Comparison proves that a limit exists; it does not compute it.
Forgetting the absolute value in $\ln|x - r|$ and the $+C$ in an indefinite integral, both of which computer algebra systems also routinely drop.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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