13.08 Integration using substitution

We used the chain rule to differentiate composite functions. Let's go back for a moment to differentiating a function like $y=\left(x^2+1\right)^4$y=(x2+1)4. Since it's a function of the form $y=f\left(x\right)^n$y=f(x)n, we can use the chain rule to differentiate. Doing this, we get:

$y$y	$=$=	$\left(x^2+1\right)^4$(x2+1)4		The function is of the form $y=f\left(x\right)^n$y=f(x)n
$\frac{dy}{dx}$dydx​	$=$=	$4\left(x^2+1\right)^3\times2x$4(x2+1)3×2x		The derivative is of the form $\frac{dy}{dx}=n\times f\left(x\right)^{n-1}\times f'\left(x\right)$dydx​=n×f(x)n−1×f′(x)
$\frac{dy}{dx}$dydx​	$=$=	$8x\left(x^2+1\right)^3$8x(x2+1)3

 

Integration by substitution is a method we can use to reverse this differentiation method. Notice that $\frac{dy}{dx}$dydx​ is of the form $f'\left(x\right)\times f\left(x\right)^n$f′(x)×f(x)n.

Integration by substitution will help us integrate functions like this.

Indefinite integrals using substitution

When we are given a substitution to use to determine an integral, it is useful to follow the following steps to arrive at a solution:

1. Rewrite the integral in the form $\int k\times f'\left(x\right)\times f\left(x\right)\ dx$∫k×f′(x)×f(x) dx
2. Using $\frac{du}{dx}$dudx​, determine a relationship between $dx$dx and $du$du 
3. Rewrite the integral in terms of $u$u and $du$du by making appropriate substitutions
4. Solve the new integral in terms of $u$u
5. Make a substitution for $u$u to rewrite the solution to the integral in terms of $x$x

Worked example

Example 1

Evaluate $\int10x\left(2+x^2\right)^4dx$∫10x(2+x2)4dx using the substitution $u=2+x^2$u=2+x2.

Think: We are given the substitution $u=f\left(x\right)=2+x^2$u=f(x)=2+x2. The first step will be to find $f'\left(x\right)$f′(x) so that we can rewrite the integral in the form:

$\int k\times f'\left(x\right)\times f\left(x\right)\ dx$∫k×f′(x)×f(x) dx

Do: Taking the derivative of $f\left(x\right)$f(x)  gives:

$f'\left(x\right)$f′(x)

$=$=

$2x$2x

 

We have $f\left(x\right)=2+x^2$f(x)=2+x2 and $f'\left(x\right)=2x$f′(x)=2x. But we have a $10x$10x term in the original integral. Therefore:

$k\times f'\left(x\right)$k×f′(x)	$=$=	$10x$10x
$k\times2x$k×2x	$=$=	$10x$10x
$k$k	$=$=	$5$5

 

Rewriting the integral in the form $\int k\times f'\left(x\right)\times f\left(x\right)\ dx$∫k×f′(x)×f(x) dx with $k=5$k=5:

$\int10x\left(2+x^2\right)^4dx$∫10x(2+x2)4dx

$=$=

$\int5\times2x\left(2+x^2\right)^4dx$∫5×2x(2+x2)4dx

 

The next step is to determine a relationship between $dx$dx and $du$du. We have:

$f'\left(x\right)=\frac{du}{dx}$f′(x)=dudx​

$=$=

$2x$2x

 

Therefore:

$du$du

$=$=

$2xdx$2xdx

 

So $2xdx$2xdx can be replaced by $du$du and $u=2+x^2$u=2+x2. Rewriting the integral in terms of $u$u gives:

$\int10x\left(2+x^2\right)^4dx$∫10x(2+x2)4dx

$=$=

$\int5u^4du$∫5u4du

 

Evaluating this integral:

$\int5u^4du$∫5u4du	$=$=	$\frac{5u^5}{5}+C$5u55​+C
$\int5u^4du$∫5u4du	$=$=	$u^5+C$u5+C

 

But $u=2+x^2$u=2+x2. Therefore, making a substitution for $u$u in terms of $x$x reveals the solution to the original integral:

$\int10x\left(2+x^2\right)^4dx$∫10x(2+x2)4dx

$=$=

$\left(2+x^2\right)^5+C$(2+x2)5+C

 

Practice questions

Question 1

Question 2

Definite integrals using substitution

Definite integrals are approached in the same manner as indefinite integrals except they require the extra step of recalculation of the limits of the integral according to the substitution variable. This also means that a final substitution is not required as the integral is evaluated in terms of the equivalent substituted variable. We will revisit our example above with integral limits added to see the difference.

Worked examples

Example 2

Evaluate $\int_0^110x\left(2+x^2\right)^4dx$∫10​10x(2+x2)4dx using the substitution $u=2+x^2$u=2+x2.

Think: We need to change the limits to $u$u values. Using $u=2+x^2$u=2+x2.

Do:

When $x=0$x=0, $u=2$u=2

When $x=1$x=1, $u=3$u=3

Using the newly calculated limits for $u$u and rewriting the integral in terms of $u$u with the substitutions determined in the worked example above gives:

$\int_0^110x\left(2+x^2\right)^4dx$∫10​10x(2+x2)4dx

$=$=

$\int_2^35u^4du$∫32​5u4du

 

Evaluating $\int_2^35u^4du$∫32​5u4du:

$\int_2^35u^4du$∫32​5u4du	$=$=	$\left[\frac{5u^5}{5}\right]_2^3$[5u55​]32​
$\int_2^35u^4du$∫32​5u4du	$=$=	$\left[u^5\right]_2^3$[u5]32​
$\int_2^35u^4du$∫32​5u4du	$=$=	$3^5-2^5$35−25
$\int_2^35u^4du$∫32​5u4du	$=$=	$243-32$243−32
$\int_2^35u^4du$∫32​5u4du	$=$=	$211$211

 

Example 3

Evaluate $\int_1^6\frac{2}{\sqrt{x+3}}dx$∫61​2√x+3​dx using the substitution $u=x+3$u=x+3.

Think: We start by calculating the new limit values for $u$u:

Do: 

Step 1: Changing the limits to $u$u values. Using $u=x+3$u=x+3:

When $x=1$x=1, $u=4$u=4

When $x=6$x=6, $u=9$u=9

Step 2: We need to replace $dx$dx by creating a relationship between $dx$dx and $du$du. The derivative of $u=x+3$u=x+3 gives:

$\frac{du}{dx}$dudx​

$=$=

$1$1

 

Rearranging this, we get:

$du$du

$=$=

$dx$dx

So $dx$dx can be replaced with $du$du.

Step 3: We rewrite the integral in terms of $u$u:

$\int_1^6\frac{2}{\sqrt{x+3}}dx$∫61​2√x+3​dx

$=$=

$\int_4^9\frac{2}{\sqrt{u}}du$∫94​2√u​du

 

Step 4: We evaluate the new integral in terms of $u$u.

$\int_4^9\frac{2}{\sqrt{u}}du$∫94​2√u​du	$=$=	$\int_4^9\frac{2}{u^{\frac{1}{2}}}du$∫94​2u12​​du
	$=$=	$\int_4^92u^{-\frac{1}{2}}du$∫94​2u−12​du
	$=$=	$\left[\frac{2u^{\frac{1}{2}}}{\frac{1}{2}}\right]_4^9$[2u12​12​​]94​
	$=$=	$\left[4u^{\frac{1}{2}}\right]_4^9$[4u12​]94​
	$=$=	$4\times9^{\frac{1}{2}}-4\times4^{\frac{1}{2}}$4×912​−4×412​
	$=$=	$4\times3-4\times2$4×3−4×2
	$=$=	$4$4

 

Practice questions

Question 3

Question 4

QUESTION 5