Integration By Substitution

Integration by substitution (also called u-substitution) is the integration counterpart of the Chain Rule. It transforms a complicated integral into a simpler one by introducing a new variable.

Step	Action
1	Choose $u = g (x)$
2	Compute du = g'(x)\,dx
3	Substitute $u$ and $d u$ ; all $x$ 's must disappear
4	Evaluate the integral in $u$
5	Replace $u$ by $g (x)$ to express the answer in $x$

The table below collects the new integration formulas established in this section. Each one is derived from a substitution in the examples that follow.

Formula	Name
$\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C$	Linear Substitution Rule
$\int \sin a x d x = - \frac{1}{a} \cos a x + C$	Sine of a Multiple
$\int \cos a x d x = \frac{1}{a} \sin a x + C$	Cosine of a Multiple
$\int \tan x d x = \ln \| \sec x \| + C$	Tangent Integral
$\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \arctan \frac{x}{a} + C$	General Arctangent
$\int a^{x} d x = \frac{a^{x}}{\ln a} + C$	General Exponential
$\int e^{a x} d x = \frac{1}{a} e^{a x} + C$	Exponential of a Multiple
$\int \sinh x d x = \cosh x + C$	Hyperbolic Sine
$\int \cosh x d x = \sinh x + C$	Hyperbolic Cosine
$\int \tanh x d x = \ln (\cosh x) + C$	Hyperbolic Tangent

The Substitution Idea

Suppose we want to evaluate an integral of the form

\int f(g(x))\,g'(x)\,dx. \tag{a}

To this end, let $g (x) = u$ . Finding the differential of both sides, we have

g'(x)\,dx = du. \tag{b}

If we substitute (b) into (a), we will have

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{ \int f(\underbrace{g(x)}_{u})\underbrace{g'(x)\,dx}_{du} = \int f(u)\,du.} \tag{c}

This equation supplies us with a method for determining integrals in a large number of cases in which the form of the integral is not obvious.

In words, the rule says: if the integrand is a composite function multiplied by the derivative of its inner function, then replacing the inner function by a single letter turns the integral into a simpler one.

The Substitution Rule

Suppose $g (x)$ is a differentiable function whose range is an interval $I$ , and $f$ is a function defined on $I$ . Also suppose $F (x) = \int f (x) d x$ for $x \in I$ . Then if $u = g (x)$ , we have

\begin{aligned} \int f(g(x))\,g'(x)\,dx &= \int f(u)\,du \\ &= F(u)+C \\ &= F(g(x))+C. \end{aligned}

Proof

Because F' = f, it follows from the chain rule that

\begin{aligned} \frac{d}{dx}F(g(x)) &= F'(g(x))\,g'(x) \\ &= f(g(x))\,g'(x). \end{aligned}

It follows from the definition of antiderivative (or integral) that

\begin{aligned} \int f(g(x))\,g'(x)\,dx &= F(g(x))+C \\ &= F(u)+C & (u = g(x)) \\ &= \int F'(u)\,du \\ &= \int f(u)\,du. \end{aligned}

Because traditionally the letter $u$ is used in the Substitution Rule, it is sometimes called u-substitution, but instead of $u$ , we can use any letter such as $t$ , $w$ , $θ$ , etc.

The u-Substitution Algorithm

Choose 
  u
, say put 
  g
  (
  x
  )
  =
  u
.
Compute du = g'(x)\,dx.
Substitute 
  u
 and 
  d
  u
 in the integral. At this step everything should be expressed in terms of 
  u
 and no 
  x
 should be present in the integral. If it is not possible, make another choice for 
  u
.
Evaluate the resulting integral.
Replace 
  u
 by 
  g
  (
  x
  )
 and express the final result in terms of 
  x
.

First Examples

The three examples below are the basic pattern: spot an inner function whose derivative is already present in the integrand, up to a constant factor.

Evaluate $\int (x^{2} + 3)^{17} x d x$ .

Solution

Let $u = x^{2} + 3$ , thus $d u = 2 x d x$ , or $\frac{1}{2} d u = x d x$ . Now we can rewrite the integral as

\begin{aligned} \int\underbrace{(x^2+3)^{17}}_{u^{17}}\underbrace{x\,dx}_{\frac{1}{2}\,du} &= \int\frac{1}{2}u^{17}\,du \\ &= \frac{1}{2}\frac{u^{18}}{18}+C \\ &= \frac{1}{36}(x^2+3)^{18}+C. \end{aligned}

Evaluate $\int x^{2} \sqrt{x^{3} + 1} d x$ .

Solution

Let $u = x^{3} + 1$ . Then $d u = 3 x^{2} d x$ or $x^{2} d x = \frac{1}{3} d u$ , and

\begin{aligned} \int x^2\sqrt{x^3+1}\,dx &= \int\sqrt{\underbrace{x^3+1}_{u}}\,\overbrace{x^2\,dx}^{\frac{1}{3}\,du} \\ &= \int\frac{1}{3}\sqrt{u}\,du \\ &= \frac{1}{3}\int u^{1/2}\,du \\ &= \frac{1}{3}\left(\frac{2}{3}u^{3/2}\right)+C \\ &= \frac{2}{9}(x^3+1)^{3/2}+C. \end{aligned}

Evaluate $\int \sin a x d x$ and $\int \cos a x d x$ .

Solution

Let $u = a x$ . Then $d u = a d x$ and

\begin{aligned} \int\sin ax\,dx &= \frac{1}{a}\int\sin u\,du \\ &= -\frac{1}{a}\cos u+C \\ &= -\frac{1}{a}\cos ax+C. \end{aligned}

Similarly,

\int \cos a x d x = \frac{1}{a} \sin a x + C .

The Linear Substitution Rule

A simple substitution is so useful that it is worth noting explicitly.

If $\int f (u) d u = F (u) + C$ , then

\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C,

where $a$ and $b$ are two constants.

Proof

Let $u = a x + b$ . Then $d u = a d x$ or $d x = \frac{1}{a} d u$ and

\begin{aligned} \int f(ax+b)\,dx &= \int\frac{1}{a}f(u)\,du \\ &= \frac{1}{a}F(u)+C \\ &= \frac{1}{a}F(ax+b)+C. \qquad (u = ax+b) \end{aligned}

In words: when the only change to a known integrand is that $x$ has been replaced by the linear expression $a x + b$ , the answer is the known antiderivative evaluated at $a x + b$ , divided by the coefficient a.

Examples Using the Linear Rule

Evaluate:

(a) $\int \sec^{2} (x - 1) d x$

(b) $\int \sin (2 x - π / 4) d x$ .

Solution

(a) Because $\int \sec^{2} x d x = \tan x + C$ , it follows from the above theorem that

\int \sec^{2} (x - 1) d x = \tan (x - 1) + C .

(b) Since $\int \sin x d x = - \cos x + C$ , the above theorem gives

\int \sin (2 x - π / 4) d x = - \frac{1}{2} \cos (2 x - π / 4) + C .

Trigonometric Integrands

Evaluate $\int \cos^{2} x \sin x d x$ .

Solution

Let $u = \cos x$ . Then $d u = - \sin x d x$ and

\begin{aligned} \int\cos^2 x\,\sin x\,dx &= -\int u^2\,du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{1}{3}\cos^3 x+C. \end{aligned}

Evaluate $\int \tan x d x$ .

Solution

Recall that $\tan x = \sin x / \cos x$ . Let $u = \cos x$ . Then $d u = - \sin x d x$ and

\begin{aligned} \int\tan x\,dx &= \int\frac{\sin x}{\cos x}\,dx \\ &= \int\frac{-du}{u} \\ &= -\ln|u|+C \\ &= -\ln|\cos x|+C. \end{aligned}

Recall that $\ln x^{α} = α \ln x$ . So

\begin{aligned} -\ln|\cos x|+C &= \ln\left(|\cos x|^{-1}\right)+C \\ &= \ln|\sec x|+C. \end{aligned}

Therefore, the result can also be written as $\int \tan x d x = \ln | \sec x | + C$ .

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{ \begin{aligned} \int\tan x\,dx &= \ln|\sec x|+C \\ \text{or}\quad &= -\ln|\cos x|+C \end{aligned}}

Integrals Leading to the Arctangent

The three examples here all reduce to $\int \frac{d u}{1 + u^{2}} = \arctan u + C$ , but each needs a different preliminary step: scaling, a change of variable, or completing the square.

Evaluate $\int \frac{1}{a^{2} + x^{2}} d x$ .

Solution

\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a^{2}} \int \frac{1}{1 + {(\frac{x}{a})}^{2}} d x .

Let $u = x / a$ . Then $d x = a d u$ and

\begin{aligned} \frac{1}{a^2}\int\frac{1}{1+(x/a)^2}\,dx &= \frac{1}{a^2}\int\frac{1}{1+u^2}\,a\,du \\ &= \frac{1}{a}\arctan u+C \\ &= \frac{1}{a}\arctan\frac{x}{a}+C. \end{aligned}

To evaluate $\int \frac{1}{1 + (x / a)^{2}} d x$ , we could simply use the Linear Substitution Rule.

Evaluate $\int \frac{x}{x^{4} + 1} d x$ .

Solution

Let $u = x^{2}$ . Then $d u = 2 x d x$ and

\begin{aligned} \int\frac{x}{x^4+1}\,dx &= \int\frac{1}{u^2+1}\,\overbrace{\frac{du}{2}}^{x\,dx} \\ &= \frac{1}{2}\arctan u+C \\ &= \frac{1}{2}\arctan x^2+C. \qquad (u = x^2) \end{aligned}

Evaluate $\int \frac{1}{x^{2} + 2 x + 5} d x$ . [Hint: Complete the square in the denominator.]

Solution

By completing the square in the denominator, we have

\begin{aligned} \int\frac{1}{x^2+2x+5}\,dx &= \int\frac{dx}{(x^2+2x+1)+4} \\ &= \int\frac{dx}{(x+1)^2+2^2}. \end{aligned}

Now let $u = x + 1$ . Then $d u = d x$ and

\begin{aligned} \int\frac{dx}{(x+1)^2+4} &= \int\frac{du}{u^2+4} \\ &= \frac{1}{2}\arctan\frac{u}{2}+C \\ &= \frac{1}{2}\arctan\frac{x+1}{2}+C. \end{aligned}

Exponential Integrands

Evaluate $\int a^{x} d x$ .

Solution

Let $a^{x} = e^{u}$ . Taking the natural logarithm of each side, we get

\begin{aligned} \ln(a^x) &= \ln e^u \\ x\ln a &= u. \end{aligned}

Thus $x \ln a = u \Rightarrow \ln a d x = d u$ . Now we can write

\begin{aligned} \int a^x\,dx &= \int e^u\frac{du}{\ln a} \\ &= \frac{1}{\ln a}e^u+C \\ &= \frac{1}{\ln a}a^x+C. \end{aligned}

Evaluate $\int e^{a x} d x$ .

Solution

We can directly use the Linear Substitution Rule or use the substitution $u = a x$ . The substitution leads to $d u = a d x$ and

\begin{aligned} \int e^{ax}\,dx &= \int e^u\frac{du}{a} \\ &= \frac{1}{a}\int e^u\,du \\ &= \frac{1}{a}e^u+C \\ &= \frac{1}{a}e^{ax}+C. \end{aligned}

Evaluate $\int \frac{e^{x}}{1 + e^{2 x}} d x$ .

Solution

Note that the substitution $u = 1 + e^{2 x}$ does not work because $d u = 2 e^{2 x} d x$ (which is not $e^{x} d x$ ) does not appear in the numerator. But if we notice that $1 + e^{2 x} = 1 + (e^{x})^{2}$ , we can make the substitution $u = e^{x}$ to make the denominator of the form $1 + u^{2}$ . Then

u = e^{x} \Rightarrow d u = e^{x} d x,

and

\begin{aligned} \int\frac{e^x}{1+e^{2x}}\,dx &= \int\frac{du}{1+u^2} \\ &= \arctan u+C \\ &= \arctan(e^x)+C. \qquad (u = e^x) \end{aligned}

Evaluate $\int \frac{\sqrt[x]{7}}{x^{2}} d x$ .

Solution

Note that $\sqrt[x]{7} = 7^{1 / x}$ . If we let $\frac{1}{x} = u$ , because $\frac{1}{x} = u \Rightarrow - \frac{1}{x^{2}} d x = d u$ , we have

\begin{aligned} \int\frac{\sqrt[x]{7}}{x^2}\,dx &= -\int 7^u\,du \\ &= -\frac{1}{\ln 7}\,7^u+C \\ &= -\frac{1}{\ln 7}\,\sqrt[x]{7}+C. \end{aligned}

Hyperbolic Integrands

Evaluate $\int \sinh x d x$ .

Solution

Because (\cosh x)' = \sinh x, we know that $\int \sinh x d x = \cosh x + C$ . We can also use the definition of $\sinh x$ :

\sinh x = \frac{e^{x} - e^{- x}}{2} .

[Recall that $\sinh x$ (similar to $\sin x$ ) is an odd function, so it is half of $e^{x} - e^{- x}$ , and $\cosh x$ (similar to $\cos x$ ) is an even function, so it is half of $e^{x} + e^{- x}$ .]

\begin{aligned} \int\sinh x\,dx &= \frac{1}{2}\int(e^x - e^{-x})\,dx \\ &= \frac{1}{2}\left[e^x - (-e^{-x})\right]+C \\ &= \cosh x+C. \end{aligned}

Similarly we can show $\int \cosh x d x = \sinh x + C$ .

Because (\cosh x)' = \sinh x, we know that $\int \sinh x d x = \cosh x + C$ . We can also use the definition of $\sinh x$ :

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{ \begin{aligned} \int\sinh x\,dx &= \cosh x+C \\ \int\cosh x\,dx &= \sinh x+C \end{aligned}}

Evaluate $\int \tanh x d x$ .

Solution

\int \tanh x d x = \int \frac{\sinh x}{\cosh x} d x .

Let $u = \cosh x$ . Then $d u = \sinh x d x$ and

\begin{aligned} \int\frac{\sinh x}{\cosh x}\,dx &= \int\frac{du}{u} \\ &= \ln|u|+C \\ &= \ln|\cosh x|+C. \end{aligned}

Because $\cosh x = (e^{x} + e^{- x}) / 2 > 0$ , we have $| \cosh x | = \cosh x$ and

\int \tanh x d x = \ln (\cosh x) + C .

Logarithmic Integrands

In both examples below the substitution is $u = \ln x$ , which works because $d u = \frac{d x}{x}$ and the factor $\frac{1}{x}$ is already present.

Evaluate $\int \frac{d x}{x \ln x}$ .

Solution

Let $u = \ln x$ , then $d u = \frac{1}{x} d x$ and

\begin{aligned} \int\underbrace{\frac{1}{\ln x}}_{1/u}\underbrace{\frac{1}{x}\,dx}_{du} &= \int\frac{du}{u} \\ &= \ln|u|+C \\ &= \ln|\ln x|+C. \qquad (u = \ln x) \end{aligned}

Evaluate $\int \frac{\sin (\ln x)}{x} d x$ .

Solution

Let $u = \ln x$ , then $d u = \frac{1}{x} d x$ . So

\begin{aligned} \int\frac{\sin(\ln x)}{x}\,dx &= \int\sin u\,du \\ &= -\cos u+C \\ &= -\cos(\ln x)+C. \end{aligned}

A Substitution Inside a Substitution

Evaluate $\int \frac{x}{1 + x^{4}} \arctan x^{2} d x$ .

Solution

Let $u = \arctan x^{2}$ , then

\begin{aligned} du &= 2x\frac{1}{1+(x^2)^2}\,dx \\ &= \frac{2x}{1+x^4}\,dx. \end{aligned}

Therefore

\begin{aligned} \int\frac{x}{1+x^4}\arctan x^2\,dx &= \int\frac{1}{2}\underbrace{\arctan x^2}_{u}\underbrace{\frac{2x}{1+x^4}\,dx}_{du} \\ &= \frac{1}{4}u^2+C \\ &= \frac{1}{4}\arctan^2 x^2+C. \end{aligned}

Exercises

Evaluate using a substitution of the form $u = g (x)$ :

(a) $\int (3 x + 2)^{5} d x$

(b) $\int x {(x^{2} - 4)}^{7} d x$

(c) $\int \frac{2 x + 1}{{(x^{2} + x + 3)}^{3}} d x$

(d) $\int x^{2} {(x^{3} - 5)}^{10} d x$

Answer

(a) $\frac{(3 x + 2)^{6}}{18} + C$

(b) $\frac{{(x^{2} - 4)}^{8}}{16} + C$

(c) $- \frac{1}{2 {(x^{2} + x + 3)}^{2}} + C$

(d) $\frac{{(x^{3} - 5)}^{11}}{33} + C$

Solution

In each part, choose $u$ to be the expression inside the power, then check that its derivative is present in the integrand up to a constant factor.

(a) Let

u = 3 x + 2

, so

d u = 3 d x

and

d x = \frac{1}{3} d u

: \begin{aligned} \int(3x+2)^{5}\,dx &= \frac{1}{3}\int u^{5}\,du \\ &= \frac{1}{3}\cdot\frac{u^{6}}{6}+C \\ &= \frac{(3x+2)^{6}}{18}+C. \end{aligned}

This is the Linear Substitution Rule in action.

(b) Let

u = x^{2} - 4

, so

d u = 2 x d x

and

x d x = \frac{1}{2} d u

: \begin{aligned} \int x\left(x^{2}-4\right)^{7}dx &= \frac{1}{2}\int u^{7}\,du \\ &= \frac{u^{8}}{16}+C \\ &= \frac{\left(x^{2}-4\right)^{8}}{16}+C. \end{aligned}
(c) Let

u = x^{2} + x + 3

. Then

d u = (2 x + 1) d x

, which is exactly the numerator: \begin{aligned} \int\frac{2x+1}{\left(x^{2}+x+3\right)^{3}}\,dx &= \int u^{-3}\,du \\ &= \frac{u^{-2}}{-2}+C \\ &= -\frac{1}{2\left(x^{2}+x+3\right)^{2}}+C. \end{aligned}
(d) Let

u = x^{3} - 5

, so

d u = 3 x^{2} d x

and

x^{2} d x = \frac{1}{3} d u

: \begin{aligned} \int x^{2}\left(x^{3}-5\right)^{10}dx &= \frac{1}{3}\int u^{10}\,du \\ &= \frac{u^{11}}{33}+C \\ &= \frac{\left(x^{3}-5\right)^{11}}{33}+C. \end{aligned}

Evaluate the following integrals involving radicals:

(a) $\int \frac{x}{\sqrt{x^{2} + 9}} d x$

(b) $\int \sqrt{4 x + 3} d x$

(c) $\int x^{3} \sqrt{x^{4} + 2} d x$

(d) $\int \frac{d x}{\sqrt{x} {(1 + \sqrt{x})}^{2}}$

Answer

(a) $\sqrt{x^{2} + 9} + C$

(b) $\frac{1}{6} (4 x + 3)^{3 / 2} + C$

(c) $\frac{1}{6} {(x^{4} + 2)}^{3 / 2} + C$

(d) $- \frac{2}{1 + \sqrt{x}} + C$

Solution

(a) Let

u = x^{2} + 9

, so

x d x = \frac{1}{2} d u

: \begin{aligned} \int\frac{x\,dx}{\sqrt{x^{2}+9}} &= \frac{1}{2}\int u^{-1/2}\,du \\ &= \frac{1}{2}\cdot 2u^{1/2}+C \\ &= \sqrt{x^{2}+9}+C. \end{aligned}
(b) Let

u = 4 x + 3

, so

d x = \frac{1}{4} d u

: \begin{aligned} \int\sqrt{4x+3}\,dx &= \frac{1}{4}\int u^{1/2}\,du \\ &= \frac{1}{4}\cdot\frac{2}{3}u^{3/2}+C \\ &= \frac{1}{6}(4x+3)^{3/2}+C. \end{aligned}
(c) Let

u = x^{4} + 2

, so

x^{3} d x = \frac{1}{4} d u

: \begin{aligned} \int x^{3}\sqrt{x^{4}+2}\,dx &= \frac{1}{4}\int u^{1/2}\,du \\ &= \frac{1}{6}u^{3/2}+C \\ &= \frac{1}{6}\left(x^{4}+2\right)^{3/2}+C. \end{aligned}
(d) Here the useful choice is

u = 1 + \sqrt{x}

, because

d u = \frac{d x}{2 \sqrt{x}}

, so

\frac{d x}{\sqrt{x}} = 2 d u

: \begin{aligned} \int\frac{dx}{\sqrt{x}\left(1+\sqrt{x}\right)^{2}} &= 2\int u^{-2}\,du \\ &= -\frac{2}{u}+C \\ &= -\frac{2}{1+\sqrt{x}}+C. \end{aligned}

Evaluate directly with the Linear Substitution Rule, without introducing $u$ :

(a) $\int \cos (5 x - 2) d x$

(b) $\int e^{3 x + 1} d x$

(c) $\int \frac{d x}{2 x + 7}$

(d) $\int \sec^{2} (1 - 4 x) d x$

(e) $\int \frac{d x}{(5 x - 1)^{3}}$

Answer

(a) $\frac{1}{5} \sin (5 x - 2) + C$

(b) $\frac{1}{3} e^{3 x + 1} + C$

(c) $\frac{1}{2} \ln | 2 x + 7 | + C$

(d) $- \frac{1}{4} \tan (1 - 4 x) + C$

(e) $- \frac{1}{10 (5 x - 1)^{2}} + C$

Solution

The rule states that if $\int f (u) d u = F (u) + C$ , then

\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C .

So in each part, write down the antiderivative you already know and divide by the coefficient of $x$ .

(a)

\int \cos x d x = \sin x + C

and

a = 5

, so the answer is

\frac{1}{5} \sin (5 x - 2) + C

.
(b)

\int e^{x} d x = e^{x} + C

and

a = 3

, so the answer is

\frac{1}{3} e^{3 x + 1} + C

.
(c)

\int \frac{d x}{x} = \ln | x | + C

and

a = 2

, so the answer is

\frac{1}{2} \ln | 2 x + 7 | + C

.
(d)

\int \sec^{2} x d x = \tan x + C

and here

a = - 4

, so the answer is

- \frac{1}{4} \tan (1 - 4 x) + C

. The negative coefficient is easy to miss.
(e)

\int x^{- 3} d x = - \frac{1}{2 x^{2}} + C

and

a = 5

, so \begin{aligned} \int\frac{dx}{(5x-1)^{3}} &= \frac{1}{5}\left(-\frac{1}{2(5x-1)^{2}}\right)+C \\ &= -\frac{1}{10(5x-1)^{2}}+C. \end{aligned}

Following the worked example for $\int \tan x d x$ , show that

\int \cot x d x = \ln | \sin x | + C,

and write the result in an equivalent form using $\csc x$ .

Answer

$\int \cot x d x = \ln | \sin x | + C = - \ln | \csc x | + C$

Solution

Write the cotangent as a quotient and let $u = \sin x$ , so that $d u = \cos x d x$ :

\begin{aligned} \int\cot x\,dx &= \int\frac{\cos x}{\sin x}\,dx \\ &= \int\frac{du}{u} \\ &= \ln|u|+C \\ &= \ln|\sin x|+C. \end{aligned}

Because $\csc x = \frac{1}{\sin x}$ , we have $| \sin x | = | \csc x |^{- 1}$ , and using $\ln x^{α} = α \ln x$ ,

\ln | \sin x | = \ln (| \csc x |^{- 1}) = - \ln | \csc x | .

Compare this with the tangent result: there the substitution was $u = \cos x$ and the derivative brought in a minus sign, which is why $\int \tan x d x = - \ln | \cos x | + C$ carries one and this one does not.

Evaluate the following trigonometric integrals:

(a) $\int \sin^{3} x \cos x d x$

(b) $\int \frac{\sin x}{\cos^{3} x} d x$

(c) $\int \tan^{2} x \sec^{2} x d x$

(d) $\int (1 + \tan θ)^{2} \sec^{2} θ d θ$

Answer

(a) $\frac{\sin^{4} x}{4} + C$

(b) $\frac{1}{2} \sec^{2} x + C$

(c) $\frac{\tan^{3} x}{3} + C$

(d) $\frac{(1 + \tan θ)^{3}}{3} + C$

Solution

In each part one trigonometric function appears to a power while its derivative appears as a separate factor. That factor is the $d u$ .

(a) Let

u = \sin x

, so

d u = \cos x d x

\int u^{3} d u = \frac{u^{4}}{4} + C = \frac{\sin^{4} x}{4} + C .

(b) Let

u = \cos x

, so

d u = - \sin x d x

: \begin{aligned} \int\frac{\sin x}{\cos^{3}x}\,dx &= -\int u^{-3}\,du \\ &= -\frac{u^{-2}}{-2}+C \\ &= \frac{1}{2\cos^{2}x}+C = \frac{1}{2}\sec^{2}x+C. \end{aligned}
(c) Let

u = \tan x

, so

d u = \sec^{2} x d x

\int u^{2} d u = \frac{u^{3}}{3} + C = \frac{\tan^{3} x}{3} + C .

(d) Let

u = 1 + \tan θ

, so

d u = \sec^{2} θ d θ

\int u^{2} d u = \frac{u^{3}}{3} + C = \frac{(1 + \tan θ)^{3}}{3} + C .

Note that $u$ need not be the inner function of a composition; here it is a whole expression whose derivative happens to be the remaining factor.

The worked example of this section established $\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \arctan \frac{x}{a} + C$ . By the same method, show that

\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \arcsin \frac{x}{a} + C (a > 0),

and use it to evaluate $\int \frac{d x}{\sqrt{9 - x^{2}}}$ .

Answer

The derivation is below; $\int \frac{d x}{\sqrt{9 - x^{2}}} = \arcsin \frac{x}{3} + C$ .

Solution

Factor $a^{2}$ out of the radical. Since $a > 0$ , $\sqrt{a^{2}} = a$ :

\begin{aligned} \int\frac{dx}{\sqrt{a^{2}-x^{2}}} &= \int\frac{dx}{\sqrt{a^{2}\left(1-\left(\frac{x}{a}\right)^{2}\right)}} \\ &= \frac{1}{a}\int\frac{dx}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}. \end{aligned}

Now let $u = \frac{x}{a}$ , so $d x = a d u$ :

\begin{aligned} \frac{1}{a}\int\frac{a\,du}{\sqrt{1-u^{2}}} &= \int\frac{du}{\sqrt{1-u^{2}}} \\ &= \arcsin u+C \\ &= \arcsin\frac{x}{a}+C. \end{aligned}

Notice the contrast with the arctangent formula: there the factor $\frac{1}{a}$ survived, here it cancels. The reason is that the radical converts $a^{2}$ into $a$ , which then cancels against the $a$ coming from $d x = a d u$ .

With $a = 3$ ,

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

Evaluate, completing the square or rescaling first:

(a) $\int \frac{d x}{x^{2} + 6 x + 13}$

(b) $\int \frac{d x}{4 x^{2} + 9}$

(c) $\int \frac{d x}{\sqrt{8 - 2 x - x^{2}}}$

Answer

(a) $\frac{1}{2} \arctan \frac{x + 3}{2} + C$

(b) $\frac{1}{6} \arctan \frac{2 x}{3} + C$

(c) $\arcsin \frac{x + 1}{3} + C$

Solution

(a) Complete the square in the denominator:

x^{2} + 6 x + 13 = (x^{2} + 6 x + 9) + 4 = (x + 3)^{2} + 2^{2} .

With $u = x + 3$ and $d u = d x$ , the general arctangent formula with $a = 2$ gives

\int \frac{d u}{u^{2} + 2^{2}} = \frac{1}{2} \arctan \frac{u}{2} + C = \frac{1}{2} \arctan \frac{x + 3}{2} + C .

(b) Factor the leading coefficient out of the denominator: \begin{aligned} \int\frac{dx}{4x^{2}+9} &= \frac{1}{4}\int\frac{dx}{x^{2}+\frac{9}{4}} \\ &= \frac{1}{4}\cdot\frac{1}{3/2}\arctan\frac{x}{3/2}+C \\ &= \frac{1}{6}\arctan\frac{2x}{3}+C. \end{aligned}
(c) Complete the square inside the radical, being careful with the sign: \begin{aligned} 8-2x-x^{2} &= 8-\left(x^{2}+2x\right) \\ &= 8-\left[(x+1)^{2}-1\right] \\ &= 9-(x+1)^{2}. \end{aligned}

With $u = x + 1$ and the arcsine formula from Exercise 6 with $a = 3$ ,

\int \frac{d u}{\sqrt{9 - u^{2}}} = \arcsin \frac{u}{3} + C = \arcsin \frac{x + 1}{3} + C .

Evaluate the following exponential integrals:

(a) $\int \frac{e^{x}}{1 + e^{x}} d x$

(b) $\int 3^{2 x} d x$

(c) $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} d x$

(d) $\int e^{x} \sqrt{1 + e^{x}} d x$

Answer

(a) $\ln (1 + e^{x}) + C$

(b) $\frac{3^{2 x}}{2 \ln 3} + C$

(c) $2 e^{\sqrt{x}} + C$

(d) $\frac{2}{3} {(1 + e^{x})}^{3 / 2} + C$

Solution

(a) Let

u = 1 + e^{x}

, so

d u = e^{x} d x

, which is exactly the numerator:

\int \frac{d u}{u} = \ln | u | + C = \ln (1 + e^{x}) + C .

The absolute value is unnecessary because $1 + e^{x} > 0$ for every $x$ .

(b) Write

3^{2 x} = 9^{x}

and apply

\int a^{x} d x = \frac{a^{x}}{\ln a} + C

with

a = 9

\int 9^{x} d x = \frac{9^{x}}{\ln 9} + C = \frac{3^{2 x}}{2 \ln 3} + C,

using $\ln 9 = \ln 3^{2} = 2 \ln 3$ . Alternatively, apply the Linear Substitution Rule to $\int 3^{u} d u$ with $u = 2 x$ .

(c) Let

u = \sqrt{x}

, so

d u = \frac{d x}{2 \sqrt{x}}

and

\frac{d x}{\sqrt{x}} = 2 d u

: \begin{aligned} \int\frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx &= 2\int e^{u}\,du \\ &= 2e^{u}+C = 2e^{\sqrt{x}}+C. \end{aligned}
(d) Let

u = 1 + e^{x}

again, so

d u = e^{x} d x

: \begin{aligned} \int e^{x}\sqrt{1+e^{x}}\,dx &= \int u^{1/2}\,du \\ &= \frac{2}{3}u^{3/2}+C \\ &= \frac{2}{3}\left(1+e^{x}\right)^{3/2}+C. \end{aligned}

Evaluate the following logarithmic integrals:

(a) $\int \frac{(\ln x)^{3}}{x} d x$

(b) $\int \frac{1 + \ln x}{x} d x$

(c) $\int \frac{d x}{x \ln x \ln (\ln x)}$

Answer

(a) $\frac{(\ln x)^{4}}{4} + C$

(b) $\frac{(1 + \ln x)^{2}}{2} + C$

(c) $\ln | \ln (\ln x) | + C$

Solution

(a) Let

u = \ln x

, so

d u = \frac{d x}{x}

\int u^{3} d u = \frac{u^{4}}{4} + C = \frac{(\ln x)^{4}}{4} + C .

(b) Let

u = 1 + \ln x

, so again

d u = \frac{d x}{x}

\int u d u = \frac{u^{2}}{2} + C = \frac{(1 + \ln x)^{2}}{2} + C .

Splitting the fraction instead gives $\ln x + \frac{(\ln x)^{2}}{2} + C$ . The two answers differ by the constant $\frac{1}{2}$ , so both are correct.

(c) This is the worked example

\int \frac{d x}{x \ln x}

with one more layer. Let

u = \ln (\ln x)

. By the Chain Rule,

d u = \frac{1}{\ln x} \cdot \frac{1}{x} d x = \frac{d x}{x \ln x},

which is precisely the rest of the integrand. Hence

\int \frac{d u}{u} = \ln | u | + C = \ln | \ln (\ln x) | + C .

Evaluate the following hyperbolic integrals:

(a) $\int \coth x d x$

(b) $\int \sinh^{2} x \cosh x d x$

(c) $\int \cosh (2 x - 1) d x$

Answer

(a) $\ln | \sinh x | + C$

(b) $\frac{\sinh^{3} x}{3} + C$

(c) $\frac{1}{2} \sinh (2 x - 1) + C$

Solution

(a) Mirroring the

\tanh x

example, write

\coth x = \frac{\cosh x}{\sinh x}

and let

u = \sinh x

, so

d u = \cosh x d x

\int \frac{d u}{u} = \ln | u | + C = \ln | \sinh x | + C .

Unlike $\cosh x$ , the function $\sinh x$ does change sign, so the absolute value must be kept here.

(b) Let

u = \sinh x

, so

d u = \cosh x d x

\int u^{2} d u = \frac{u^{3}}{3} + C = \frac{\sinh^{3} x}{3} + C .

(c) Since

\int \cosh x d x = \sinh x + C

, the Linear Substitution Rule with

a = 2

gives

\int \cosh (2 x - 1) d x = \frac{1}{2} \sinh (2 x - 1) + C .

Sometimes the substitution must be inverted and back-substituted. Evaluate:

(a) $\int \frac{x}{x + 1} d x$

(b) $\int \frac{x^{2}}{x - 1} d x$

(c) $\int x \sqrt{x - 1} d x$

Answer

(a) $x - \ln | x + 1 | + C$

(b) $\frac{(x - 1)^{2}}{2} + 2 (x - 1) + \ln | x - 1 | + C$

(c) $\frac{2}{5} (x - 1)^{5 / 2} + \frac{2}{3} (x - 1)^{3 / 2} + C$

Solution

In each part the derivative of the inner function is not sitting in the integrand. The fix is to solve the substitution for $x$ and replace every remaining $x$ as well.

(a) Let

u = x + 1

, so

x = u - 1

and

d x = d u

: \begin{aligned} \int\frac{x}{x+1}\,dx &= \int\frac{u-1}{u}\,du \\ &= \int\left(1-\frac{1}{u}\right)du \\ &= u-\ln|u|+C \\ &= x+1-\ln|x+1|+C. \end{aligned}

Absorbing the constant $1$ into $C$ gives $x - \ln | x + 1 | + C$ .

(b) Let

u = x - 1

, so

x = u + 1

and

x^{2} = (u + 1)^{2} = u^{2} + 2 u + 1

: \begin{aligned} \int\frac{x^{2}}{x-1}\,dx &= \int\frac{u^{2}+2u+1}{u}\,du \\ &= \int\left(u+2+\frac{1}{u}\right)du \\ &= \frac{u^{2}}{2}+2u+\ln|u|+C \\ &= \frac{(x-1)^{2}}{2}+2(x-1)+\ln|x-1|+C. \end{aligned}
(c) Let

u = x - 1

, so

x = u + 1

and

d x = d u

: \begin{aligned} \int x\sqrt{x-1}\,dx &= \int(u+1)u^{1/2}\,du \\ &= \int\left(u^{3/2}+u^{1/2}\right)du \\ &= \frac{2}{5}u^{5/2}+\frac{2}{3}u^{3/2}+C \\ &= \frac{2}{5}(x-1)^{5/2}+\frac{2}{3}(x-1)^{3/2}+C. \end{aligned}

Evaluate $\int \frac{d x}{\sqrt{x} (1 + x)}$ .

Answer

$2 \arctan \sqrt{x} + C$

Solution

The key observation is that $1 + x = 1 + {(\sqrt{x})}^{2}$ , which turns the denominator into the arctangent form. Let $u = \sqrt{x}$ , so that

d u = \frac{d x}{2 \sqrt{x}} ⟹ \frac{d x}{\sqrt{x}} = 2 d u .

Then

\begin{aligned} \int\frac{dx}{\sqrt{x}\,(1+x)} &= \int\frac{2\,du}{1+u^{2}} \\ &= 2\arctan u+C \\ &= 2\arctan\sqrt{x}+C. \end{aligned}

This is the same manoeuvre used in the worked example for $\int \frac{e^{x}}{1 + e^{2 x}} d x$ : rewrite the denominator as $1 + (something)^{2}$ , then let $u$ be that something.

Evaluate $\int \sin x \cos x d x$ in three ways: with $u = \sin x$ , with $u = \cos x$ , and by first using the identity $\sin 2 x = 2 \sin x \cos x$ . Explain why the three answers look different but are all correct.

Answer

$\frac{\sin^{2} x}{2} + C$ , $- \frac{\cos^{2} x}{2} + C$ , and $- \frac{\cos 2 x}{4} + C$ . The three differ from one another by constants, so they describe the same family of antiderivatives.

Solution

Method 1. Let

u = \sin x

, so

d u = \cos x d x

\int u d u = \frac{u^{2}}{2} + C = \frac{\sin^{2} x}{2} + C .

Method 2. Let

u = \cos x

, so

d u = - \sin x d x

- \int u d u = - \frac{u^{2}}{2} + C = - \frac{\cos^{2} x}{2} + C .

Method 3. Using

\sin x \cos x = \frac{1}{2} \sin 2 x

and the formula

\int \sin a x d x = - \frac{1}{a} \cos a x + C

with

a = 2

: \begin{aligned} \int\sin x\cos x\,dx &= \frac{1}{2}\int\sin 2x\,dx \\ &= \frac{1}{2}\left(-\frac{1}{2}\cos 2x\right)+C \\ &= -\frac{\cos 2x}{4}+C. \end{aligned}
Why they agree. Using

\sin^{2} x + \cos^{2} x = 1

\frac{\sin^{2} x}{2} - (- \frac{\cos^{2} x}{2}) = \frac{\sin^{2} x + \cos^{2} x}{2} = \frac{1}{2},

and using $\cos 2 x = 1 - 2 \sin^{2} x$ ,

\frac{\sin^{2} x}{2} - (- \frac{\cos 2 x}{4}) = \frac{1}{4} .

Each pair differs by a constant, which the arbitrary constant $C$ absorbs. Two antiderivatives of the same function on an interval can differ only by a constant, so answers that look unrelated may still both be right.

Each line below contains a substitution error. Find it and give the correct result:

(a) $\int {(x^{2} + 3)}^{17} d x = \frac{{(x^{2} + 3)}^{18}}{18 \cdot 2 x} + C$

(b) $\int x \sin (x^{2}) d x$ : "let $u = x^{2}$ , then $\int \sin u d u = - \cos u + C = - \cos (x^{2}) + C$ ."

(c) $\int (2 x + 1)^{5} d x = \frac{(2 x + 1)^{6}}{6} + C$

Answer

(a) $2 x$ is not a constant and cannot be moved outside the integral. There is no elementary substitution here; the integrand must be expanded.

(b) The factor $\frac{1}{2}$ from $d u = 2 x d x$ was dropped. Correct: $- \frac{1}{2} \cos (x^{2}) + C$ .

(c) The factor $\frac{1}{2}$ from $d x = \frac{1}{2} d u$ was dropped. Correct: $\frac{(2 x + 1)^{6}}{12} + C$ .

Solution

Every one of these is caught by the same test: differentiate the proposed answer and see whether the integrand returns.

(a) The substitution

u = x^{2} + 3

requires

d u = 2 x d x

, but the integrand contains no factor of

x

. Writing

\frac{1}{2 x}

outside treats a function of

x

as a constant, which the Quotient Rule immediately exposes: differentiating the proposed answer gives

\frac{{(x^{2} + 3)}^{17} (35 x^{2} - 3)}{36 x^{2}},

not ${(x^{2} + 3)}^{17}$ . Substitution simply does not apply here. Compare this with the first worked example of the section, $\int {(x^{2} + 3)}^{17} x d x$ , where the extra factor $x$ is exactly what makes the method work.

(b) From

u = x^{2}

we get

d u = 2 x d x

, so

x d x = \frac{1}{2} d u

, not

d u

. Carrying the factor properly, \begin{aligned} \int x\sin\left(x^{2}\right)dx &= \frac{1}{2}\int\sin u\,du \\ &= -\frac{1}{2}\cos u+C \\ &= -\frac{1}{2}\cos\left(x^{2}\right)+C. \end{aligned}
(c) From

u = 2 x + 1

we get

d u = 2 d x

, so

d x = \frac{1}{2} d u

. Then \begin{aligned} \int(2x+1)^{5}dx &= \frac{1}{2}\cdot\frac{u^{6}}{6}+C \\ &= \frac{(2x+1)^{6}}{12}+C. \end{aligned}

Differentiating the student's answer gives $(2 x + 1)^{5} \cdot 2$ , which is twice the integrand. This is the single most common substitution slip.

For each integral, state whether the suggested substitution works, and say why:

(a) $\int x e^{x^{2}} d x$ with $u = x^{2}$

(b) $\int e^{x^{2}} d x$ with $u = x^{2}$

(c) $\int \frac{\ln x}{x} d x$ with $u = x$

(d) $\int x^{2} \sqrt{x^{3} + 1} d x$ with $u = x^{3} + 1$

Answer

(a) Works. Result: $\frac{1}{2} e^{x^{2}} + C$ .

(b) Fails. There is no factor of $x$ to absorb into $d u$ .

(c) Useless. It changes nothing. The right choice is $u = \ln x$ , giving $\frac{1}{2} (\ln x)^{2} + C$ .

(d) Works. Result: $\frac{2}{9} {(x^{3} + 1)}^{3 / 2} + C$ .

Solution

Step 3 of the algorithm is the test: after substituting, no $x$ may remain. If one does, the choice of $u$ was wrong.

(a) With

u = x^{2}

we get

d u = 2 x d x

, and the integrand supplies exactly the factor

x

\int x e^{x^{2}} d x = \frac{1}{2} \int e^{u} d u = \frac{1}{2} e^{x^{2}} + C .

(b) With

u = x^{2}

we would need

d x = \frac{d u}{2 x} = \frac{d u}{2 \sqrt{u}}

, leaving

\int \frac{e^{u}}{2 \sqrt{u}} d u,

which is no simpler. In fact $\int e^{x^{2}} d x$ cannot be expressed in terms of elementary functions at all, a reminder from Section 5.2 that integration does not always succeed.

(c) Setting

u = x

gives

d u = d x

and reproduces the original integral unchanged. A substitution is only useful when it simplifies the integrand. The productive choice is

u = \ln x

, since

d u = \frac{d x}{x}

\int \frac{\ln x}{x} d x = \int u d u = \frac{(\ln x)^{2}}{2} + C .

(d) With

u = x^{3} + 1

we get

d u = 3 x^{2} d x

, and the factor

x^{2}

is present. This is the second worked example of the section, with answer

\frac{2}{9} {(x^{3} + 1)}^{3 / 2} + C

A classical trick. Multiply the integrand by $\frac{\sec x + \tan x}{\sec x + \tan x}$ and use a substitution to show that

\int \sec x d x = \ln | \sec x + \tan x | + C .

Then find $\int \csc x d x$ by the analogous device.

Answer

$\int \sec x d x = \ln | \sec x + \tan x | + C$ and $\int \csc x d x = - \ln | \csc x + \cot x | + C$ .

Solution

The secant. Multiplying by a well-chosen form of $1$ ,

\begin{aligned} \int\sec x\,dx &= \int\frac{\sec x\left(\sec x+\tan x\right)}{\sec x+\tan x}\,dx \\ &= \int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}\,dx. \end{aligned}

Now let $u = \sec x + \tan x$ . Then

d u = (\sec x \tan x + \sec^{2} x) d x,

which is exactly the numerator. Hence

\begin{aligned} \int\sec x\,dx &= \int\frac{du}{u} \\ &= \ln|u|+C \\ &= \ln|\sec x+\tan x|+C. \end{aligned}

The trick works because the derivative of $\sec x + \tan x$ happens to be $\sec x$ times $\sec x + \tan x$ itself.

The cosecant. Multiply instead by $\frac{\csc x + \cot x}{\csc x + \cot x}$ and let $u = \csc x + \cot x$ , so that

\begin{aligned} du &= \left(-\csc x\cot x-\csc^{2}x\right)dx \\ &= -\csc x\left(\cot x+\csc x\right)dx. \end{aligned}

Therefore

\begin{aligned} \int\csc x\,dx &= \int\frac{\csc x\left(\csc x+\cot x\right)}{\csc x+\cot x}\,dx \\ &= -\int\frac{du}{u} \\ &= -\ln|\csc x+\cot x|+C. \end{aligned}

In each integral the substitution is an inverse function. Evaluate:

(a) $\int \frac{\arcsin x}{\sqrt{1 - x^{2}}} d x$

(b) $\int \frac{\arctan x}{1 + x^{2}} d x$

(c) $\int \frac{e^{\arctan x}}{1 + x^{2}} d x$

Answer

(a) $\frac{(\arcsin x)^{2}}{2} + C$

(b) $\frac{(\arctan x)^{2}}{2} + C$

(c) $e^{\arctan x} + C$

Solution

The pattern in all three is that the derivative of the inverse function is the other factor in the integrand.

(a) Let

u = \arcsin x

, so

d u = \frac{d x}{\sqrt{1 - x^{2}}}

\int u d u = \frac{u^{2}}{2} + C = \frac{(\arcsin x)^{2}}{2} + C .

(b) Let

u = \arctan x

, so

d u = \frac{d x}{1 + x^{2}}

\int u d u = \frac{u^{2}}{2} + C = \frac{(\arctan x)^{2}}{2} + C .

(c) With the same

u = \arctan x

\int e^{u} d u = e^{u} + C = e^{\arctan x} + C .

Compare part (b) with the last worked example of the section, where $u = \arctan x^{2}$ required the Chain Rule to produce $d u = \frac{2 x d x}{1 + x^{4}}$ .

Prove the Linear Substitution Rule in the special case $f (u) = u^{n}$ with $n \neq - 1$ : show directly that

\int (a x + b)^{n} d x = \frac{(a x + b)^{n + 1}}{a (n + 1)} + C,

and explain what goes wrong when $n = - 1$ .

Answer

Substituting $u = a x + b$ reduces the integral to $\frac{1}{a} \int u^{n} d u$ . When $n = - 1$ the power rule fails and the answer becomes $\frac{1}{a} \ln | a x + b | + C$ instead.

Solution

Let $u = a x + b$ . Then $d u = a d x$ , so $d x = \frac{1}{a} d u$ , and

\begin{aligned} \int(ax+b)^{n}dx &= \frac{1}{a}\int u^{n}\,du \\ &= \frac{1}{a}\cdot\frac{u^{n+1}}{n+1}+C \\ &= \frac{(ax+b)^{n+1}}{a(n+1)}+C, \end{aligned}

valid whenever $n \neq - 1$ .

The case $n = - 1$ . The step $\int u^{n} d u = \frac{u^{n + 1}}{n + 1}$ divides by $n + 1$ , which is zero when $n = - 1$ . This is the same exception met in the power rule for integration. Using the logarithmic form instead,

\begin{aligned} \int\frac{dx}{ax+b} &= \frac{1}{a}\int\frac{du}{u} \\ &= \frac{1}{a}\ln|u|+C \\ &= \frac{1}{a}\ln|ax+b|+C. \end{aligned}

Check. Differentiating, $\frac{d}{d x} (\frac{1}{a} \ln | a x + b |) = \frac{1}{a} \cdot \frac{a}{a x + b} = \frac{1}{a x + b}$ , as required.

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