1.1 Substitution

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In this section we reverse the chain rule by making a substitution.

$U$ -Substitution

We make a $u$ -substitution to fill in the gaps in the following equation, which reverses the chain rule:

$\int f(g(x))g'(x) \ dx = F(g(x)) + C.$

If we let $u$ be the inside of the composition and we compute $du$ :

$\begin{align*} &\text{Let} \quad \;\;u = g(x) \\ &\text{then} \quad du = g'(x) \ dx \end{align*}$ Substituting $u$ and $du$ into the original integral gives

$\int f(g(x))g'(x) \ dx = \int f(u) \ du$ which we can compute if we know an anti-derivative of $f(u)$ : $=F(u) + C,$ Finally, we can back substitute to get the final answer: $=F(g(x)) + C.$ Now we can see that $\int f(g(x))g'(x) \ dx = F(g(x)) + C.$

example 1 Compute $\int 2x\cos (x^2 + 1) \ dx.$

First, note that the function $\cos (x^2 + 1)$ is a composition. The inside function is $x^2 + 1$ . Moreover, the derivative of the inside function is $2x$ which is also present in our integrand. These observations suggest that we should try the substitution $u = x^2 + 1$ . If we let $u = x^2 + 1$ , then using the differential form of the derivative, we have $du = 2x \ dx$

Now, the original integral which was in terms of the variable $x$ and the differential $dx$ can be rewritten in terms of the new variable $u$ and its differential $du$ . Substituting $u$ for $x^2 + 1$ and $du$ for $2x \ dx$ , the original integral becomes $\int 2x\cos (x^2 + 1) \ dx = \int \cos (x^2 + 1) \ 2x\ dx = \int \cos u \ du.$ We now have a basic integral in the variable $u$ which can easily be computed: $\int \cos u \ du = \sin u + C,$ This is essentially our answer except that it is in the wrong variable. We must go back to $x$ using $u = x^2 + 1$ : $\sin u + C = \sin (x^2 + 1) + C.$ Thus, using $u$ -substitution we can conclude that $\int 2x\cos (x^2 + 1) \ dx = \sin (x^2 + 1) + C.$ As a final remark, we can check our answer by differentiation: $\frac {d}{dx} \sin (x^2 + 1) = \cos (x^2 +1) \cdot 2x \;\; \text {by the chain rule}$ This agrees with our original integrand, so our answer is correct!

(problem 1)
Compute: $\displaystyle {\int 5x^4\sin (x^5 -7) \ dx}$ .
Let $u = \answer {x^5 - 7}$ . Then $du = \answer {5x^4 dx}$ .

Don’t forget the ‘dx’ in your answer for ‘du’.

Convert to an integral in the variable $u$ : $\int 5x^4\sin (x^5 -7) \ dx = \int \answer {\sin u du}$

Don’t forget the ‘du’ in your integral.

The final answer in terms of $x$ is: $\int 5x^4\sin (x^5 -7) \ dx = \answer {-\cos (x^5 -7)} +C.$

example 2 Compute $\int 2x(x^2 + 1)^4 \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

and the integral can be written as $\int 2x(x^2 + 1)^4 \ dx = \int (x^2 + 1)^4 \ 2x \ dx = \int u^4 \ du.$ The last integral can be computed as $\int u^4 \ du = \tfrac {u^5}{5} + C,$ and by back substituting, we have $\tfrac {u^5}{5} + C = \tfrac 15(x^2 + 1)^5 + C .$ Thus, using $u$ -substitution we can conclude that $\int 2x(x^2 + 1)^4 \ dx = \tfrac 15(x^2 + 1)^5 + C.$

(problem 2)
Compute $\displaystyle {\int 3x^2(x^3 + 2)^6 \ dx}$ .
Let $u = \answer {x^3 + 2}$ . Then $du = \answer {3x^2 dx}$ .

Don’t forget the ‘dx’ in your answer for ‘du’.

Convert to an integral in the variable $u$ : $\int 3x^2(x^3 + 2)^6 \ dx = \int \answer {u^6 du}$

Don’t forget the ‘du’ in your integral.

The final answer in terms of $x$ is:

$\int 3x^2(x^3 + 2)^6 \ dx = \answer {(x^3 + 2)^7/7} +C.$

example 3 Compute $\int _0^2 2xe^{x^2 + 1} \ dx.$ In this example, we have a definite integral. When making a substitution in a definite integral, we must change the endpoints of integration in addition to the integrand and the differential. As before, let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$ Now we prepare to change the endpoints. If $x = 0$ , then $u = 0^2 + 1 = 1$ and if $x = 2$ , then $u = 2^2 +1 = 5$ . So, the endpoints in the original integral, 0 and 2, will be changed to 1 and 5 respectively. We get $\int _0^2 2xe^{x^2 + 1} \ dx = \int _1^5 e^u \ du.$ The last integral can be computed as $\int _1^5 e^u \ du = e^u \Big {|}_1^5 = e^5 - e.$

Thus, using $u$ -substitution we can conclude that $\int _0^2 2xe^{x^2 + 1} \ dx = e^5 - e.$

(problem 3)
Compute $\displaystyle {\int _1^2 -2xe^{-x^2} \ dx}$ .
Let $u = \answer {-x^2}$ . Then $du = \answer {-2x dx}$ .

Don’t forget the ‘dx’ in your answer for ‘du’.

If $x = 1$ , then $u = \answer {-1}$ and
if $x = 2$ , then $u = \answer {-4}$
Convert to a definite integral in the variable $u$ : $\int _1^2 -2xe^{-x^2} \ dx = \int _{\answer {-1}}^{\answer {-4}} \answer {e^u du}$

Don’t forget the ‘du’ in your integral.

The final answer is: $\int _1^2 -2xe^{-x^2} \ dx = \answer {e^{-4} - e^{-1}}$

example 4 Compute $\int 2x\sqrt {x^2 + 1} \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

$\int 2x\sqrt {x^2 + 1} \ dx = \int \sqrt {x^2 + 1} \cdot 2x\ dx = \int \sqrt {u} \ du.$ The last integral can be computed as $\int \sqrt u \ du = \int u^{1/2} \ du = \tfrac {u^{3/2}}{3/2} + C = \tfrac 23 u^{3/2} + C,$ and by back substituting, we have $\tfrac 23 u^{3/2} + C = \frac 23 (x^2 + 1)^{3/2} + C.$ Thus, using $u$ -substitution we can conclude that $\int 2x\sqrt {x^2 + 1} \ dx = \tfrac 23 (x^2 + 1)^{3/2} + C.$

(problem 4) Compute $\displaystyle {\int 4x^3 \sqrt {x^4 +2} \ dx}$ .
Let $\;u = \answer {x^4 + 2}$ ,
then $\;du = \answer {4x^3 dx}$

Don’t forget the ‘dx’ in your answer for ‘du’.

Convert to an integral in the variable $u$ : $\int 4x^3 \sqrt {x^4 +2} \ dx = \int \answer {\sqrt u du}$

Don’t forget the ‘du’ in your integral.

$\int 4x^3 \sqrt {x^4 +2} \ dx = \answer {2/3(x^4 +2)^{3/2}} +C.$

example 5 Compute $\int \frac {2x}{x^2 + 1} \ dx.$ Let $u = x^2 + 1$ . We have, $u = x^2 + 1$ $du = 2x \ dx$

$\int \frac {2x}{x^2 + 1} \ dx = \int \frac {1}{x^2 + 1} \cdot 2x\ dx = \int \frac {1}{u} \ du.$ The last integral can be computed as $\int \frac {1}{u} \ du = \ln |u| + C,$ and by back substituting, we have $\ln |u| + C = \ln |x^2 + 1| + C=\ln (x^2 + 1) + C.$ Thus, using $u$ -substitution we can conclude that $\int \frac {2x}{x^2 + 1} \ dx = \ln (x^2 + 1) + C.$

(problem 5) Compute $\displaystyle {\int \frac {4x^3}{x^4 +2} \ dx}$ .
Let $\;u = \answer {x^4 + 2}$ ,
then $\;du = \answer {4x^3 dx}$ .

Don’t forget the ‘dx’ in your answer for ‘du’.

Convert to an integral in the variable $u$ : $\int \frac {4x^3}{x^4 +2} \ dx = \int \answer {1/u du}$

Don’t forget the ‘du’ in your integral.

Back substitute: $\int \frac {4x^3}{x^4 +2} \ dx = \answer {\ln |x^4 +2|} +C.$

example 6 Compute $\int e^x\sin (e^x) \ dx.$ We let $u = e^x$ . We have, $u = e^x$ $du = e^x \ dx$

and the integral can be written as $\int e^x\sin (e^x) \ dx = \int \sin (e^x) \cdot e^x \ dx = \int \sin (u) \ du.$ The last integral can be computed as $\int \sin u \ du = -\cos u + C,$ and by back substituting, we have $-\cos u + C = -\cos (e^x) + C.$ Thus, using $u$ -substitution we can conclude that $\int e^x\sin (e^x) \ dx = -\cos (e^x) + C.$

(problem 6) Compute $\displaystyle {\int \frac {\cos (\ln x)}{x} \ dx}$ .
Let $\;u = \answer {\ln x}$ ,
then $\;du = \answer {1/x dx}$

Don’t forget the ‘dx’ in your answer for ‘du’

Convert to an integral in the variable $u$ : $\int \frac {\cos (\ln x)}{x} \ dx = \int \answer {\cos u du}$

Don’t forget the ‘du’ in your integral

$\int \frac {\cos (\ln x)}{x} \ dx = \answer {\sin (\ln x)} +C$

example 7 Compute $\int 3x^2\sec (x^3)\tan (x^3) \ dx.$ Let $u = x^3$ . Then we have, $u = x^3$ $du = 3x^2 \ dx$

and the integral can be written as

$\begin{align*} \int 3x^2\sec(x^3)\tan(x^3) \ dx &= \int \sec(x^3)\tan(x^3) \cdot 3x^2 \ dx\\ &= \int \sec u\tan u \ du. \end{align*}$ The last integral can be computed as $\int \sec u\tan u \ du = \sec u + C,$ and by back substituting, we have $\sec u + C = \sec (x^3) + C.$ Thus, using $u$ -substitution we can conclude that $\int 3x^2\sec (x^3)\tan (x^3) \ dx = \sec (x^3) + C.$

(problem 7)

Let $u = e^x$

Compute $du$

$\int e^x\sec (e^x)\tan (e^x) \ dx = \answer {\sec (e^x)} +C.$

example 8 Compute $\int x^3\cos (x^4) \ dx.$ Let $u = x^4$ . Then, $u = x^4$ $du = 4x^3 \ dx$

and the integral can be written as

$\begin{align*} \int x^3\cos(x^4) \ dx &= \int \cos(x^4) \cdot x^3\ dx \\ &= \int \cos(x^4)\cdot \tfrac14 \cdot 4x^3\ dx\\ &= \int \tfrac14\cos u \ du. \end{align*}$ The last integral can be computed as $\int \tfrac 14 \cos u \ du = \tfrac 14 \sin u + C,$ and by back substituting, we have $\tfrac 14 \sin u + C = \tfrac 14 \sin (x^4) + C.$ Thus, using $u$ -substitution we can conclude that $\int x^3\cos (x^4) \ dx = \tfrac 14 \sin (x^4) + C.$

(problem 8)

Let $u = x^2$

Compute $du$

$\int x\cos (x^2) \ dx = \answer {\sin (x^2)/2} +C.$

example 9 Compute $\int xe^{-x^2} \ dx.$ Let $u = -x^2$ . Then we have, $u = -x^2$ $du = -2x \ dx$ and the integral can be written as $\begin{align*} \int xe^{-x^2} \ dx &= \int e^{-x^2} \cdot x\ dx \\ &= \int e^{-x^2}( -\tfrac12)\ (-2x)\ dx \\ &= \int -\tfrac12 e^u \ du. \end{align*}$ The last integral can be computed as $\int -\tfrac 12 e^u \ du = -\tfrac 12 e^u + C,$ and by back substituting, we have $-\tfrac 12 e^u + C = -\tfrac 12 e^{-x^2} + C.$ Thus, using $u$ -substitution we can conclude that $\int xe^{-x^2} \ dx = -\tfrac 12 e^{-x^2} + C.$

(problem 9)

Let $u = x^3$

Compute $du$

$\int x^2e^{x^3} \ dx = \answer {e^{x^3}/3} +C.$

example 10 Compute $\int \sin (3x) \ dx.$ Let $u = 3x$ . Then we have, $u = 3x$ $du = 3 \ dx$ and the integral can be written as $\int \sin (3x) \ dx = \int \sin (3x) \cdot \tfrac 13\cdot 3 \ dx = \int \tfrac 13 \sin u \ du.$ The last integral can be computed as $\int \tfrac 13 \sin u \ du = -\tfrac 13 \cos u + C,$ and by back substituting, we have $-\tfrac 13 \cos u + C = -\tfrac 13 \cos (3x) + C.$ Thus, using $u$ -substitution we can conclude that $\int \sin (3x) \ dx = -\tfrac 13 \cos (3x) + C.$

(problem 10)

Let $u = 3x$

Compute $du$

$\int \cos (3x) \ dx = \answer {1/3 \sin (3x)} +C.$

example 11 Compute $\int e^{x/2} \ dx.$ Let $u = \frac {x}{2}$ , then $du = \tfrac 12 \ dx$ . Since $2 \cdot \frac 12 = 1$ , the integral can be written as $\int e^{x/2} \ dx = 2\int \tfrac {1}{2}e^{x/2} \ dx = 2 \int e^u \ du.$ The last integral can be computed as $2\int e^u \ du = 2 e^u + C,$ and by back substituting, we have $2e^u + C = 2e^{x/2}+ C.$ Thus, using $u$ -substitution we can conclude that $\int e^{x/2} \ dx = 2e^{x/2} + C.$

(problem 11)

Let $u = 5x$

Compute $du$

$\int e^{5x} \ dx = \answer {1/5 e^{5x}} +C.$

example 12 Compute $\int (4x+3)^5 \ dx.$ Let $u = 4x+3$ . Then we have, $u = 4x+3$ $du = 4 \ dx$ and the integral can be written as $\int (4x+3)^5 \ dx = \int (4x+3)^5 \cdot \tfrac 14\cdot 4 \ dx = \int \tfrac 14 u^5 \ du.$ The last integral can be computed as $\tfrac 14 \int u^5 \ du = \tfrac 14 \cdot \tfrac {u^6}{6} + C = \tfrac {u^6}{24} + C,$ and by back substituting, we have $\tfrac {u^6}{24} + C = \tfrac {(4x+3)^6}{24}+ C.$ Thus, using $u$ -substitution we can conclude that $\int (4x+3)^5 \ dx = \tfrac {1}{24}(4x+3)^6 + C.$

(problem 12)

Let $u = 2x - 5$

Compute $du$

$\int (2x - 5)^4 \ dx = \answer {\frac {1}{10}(2x-5)^5} +C.$

example 13 Compute $\dis \int \frac {1}{x\ln ^2 x} \ dx.$
Step 1) Transform the integral: $\begin{align*} \int \frac{1}{x\ln^2 x}& \ dx \, = \, \int \frac{1}{\ln^2 x} \cdot \frac{1}{x} \, dx\, = \, \int \frac{1}{u^2} \ du\\ \text{let} \;\; u = & \ \ln(x)& \\ \text{then} \;\; du = & \ \frac{1}{x} \ dx& \end{align*}$ Step 2) Anti-differentiate: $\int \frac {1}{u^2} \, du \, = \, \int u^{-2} \, du\, =\, \frac {u^{-1}}{-1} + C\, =\, -\frac {1}{u} + C$ Step 3) Back substitute: $-\frac {1}{u} + C \, =\, -\frac {1}{\ln x} + C$ Hence, $\int \frac {1}{x\ln ^2 x} \ dx \,= \,-\frac {1}{\ln x} + C$

Let $u = \ln x$ . Then we have, $u = \ln x$ $du = \tfrac 1x \ dx$ and the integral can be written as $\int \frac {1}{x\ln ^2 x} \ dx = \int \frac {1}{\ln ^2 x} \cdot \frac {1}{x}\ dx = \int \frac {1}{u^2} \ du.$ The last integral can be computed as $\int \frac {1}{u^2} \ du = \int u^{-2} \ du = \tfrac {u^{-1}}{-1} + C = -\frac {1}{u} + C,$ and by back-substituting, we have $-\frac {1}{u} + C = -\frac {1}{\ln x} + C.$ Thus, using $u$ -substitution we can conclude that $\int \frac {1}{x\ln ^2 x} \ dx = -\frac {1}{\ln x} + C.$

(problem 13)

Let $u = \ln x$

Compute $du$

$\int \frac {1}{x\ln ^3 x} \ dx = \answer {- \ln ^{-2} x/2} +C.$

example 14 Compute $\int \sin ^4 x\cos x \ dx.$ Let $u = \sin x$ . Then we have, $u = \sin x$ $du = \cos x \ dx$ and the integral can be written as $\int \sin ^4 x\cos x \ dx = \int u^4 \ du.$ The last integral can be computed as $\int u^4 \ du = \tfrac {u^5}{5} + C,$ and by back-substituting, we have $\tfrac {u^5}{5} + C = \tfrac 15 \sin ^5 x + C.$ Thus, using $u$ -substitution we can conclude that $\int \sin ^4 x\cos x \ dx = \tfrac 15 \sin ^5 x + C.$

(problem 14)

Let $u = \tan x$

Compute $du$

$\int \tan ^5 x\sec ^2 x \ dx = \answer {\tan ^6 x/6} +C.$

example 15 Compute $\int \tan x \ dx.$ First rewrite the integral: $\int \tan x \ dx =\int \frac {\sin x}{\cos x} \ dx.$ Now, let $u = \cos x$ ; then $du = -\sin x \ dx$ and the integral can be written as $\begin{align*} \int \frac{\sin x}{\cos x} \ dx &= \int \frac{1}{\cos x}\ \sin x \ dx \\ &= - \int \frac{1}{\cos x}\ \big(-\sin(x)\big) \ dx\\ &=-\int \frac{1}{u} \ du. \end{align*}$ The last integral can be computed as $-\int \frac {1}{u} \ du = -\ln |u| + C,$ and by back-substituting, we have $-\ln |u| + C = -\ln |\cos x| + C = \ln |\sec x| +C.$ Thus, using $u$ -substitution we can conclude that $\int \tan x \ dx = \ln |\sec x| + C.$

(problem 15)

Rewrite: $\cot x = \frac {\cos x}{\sin x}$

Let $u = \sin x$

Compute $du$

$\int \cot x \ dx = \answer {\ln |\sin (x)|} +C.$

example 16 Compute $\int \sec x \ dx.$ First rewrite the integral: $\int \sec x \ dx =\int \sec x\frac {\sec x+\tan x}{\sec x+\tan x} \ dx.$ Distributing in the numerator, we get $\int \sec x\frac {\sec x+\tan x}{\sec x+\tan x} \ dx = \int \frac {\sec ^2 x+\sec x\tan x}{\sec x+\tan x} \ dx.$ Now, let $u = \sec x + \tan x$ . Then $du = [\sec x\tan x + \sec ^2 x] \ dx$ and the integral can be rewritten as $\int \frac {\sec ^2 x+\sec x\tan x}{\sec x+\tan x} \ dx = \int \frac {1}{u} \ du.$ The last integral can be computed as $\int \frac {1}{u} \ du = \ln |u| + C,$ and by back-substituting, we have $\ln |u| + C = \ln |\sec x + \tan x| + C.$ Thus, using $u$ -substitution we can conclude that $\int \sec x \ dx = \ln |\sec x + \tan x| + C.$

Here is a detailed, lecture style video on $u$ -substitution:

U-Substitution

$U$ -Substitution