Integration by Substitution

$\int \limits _{x=a}^{x=b}f(x)dx$	$=\int \limits _{x=a}^{x=b}f(x)\ {\frac {du}{du}}\ dx$	(1)	i.e. ${\frac {du}{du}}=1$
	$=\int \limits _{x=a}^{x=b}{\left(f(x)\ {\frac {dx}{du}}\right)\left({\frac {du}{dx}}\right)}\ dx$	(2)	i.e. ${\frac {dx}{du}}\cdot {\frac {du}{dx}}=1$
	$=\int \limits _{x=a}^{x=b}\left(f(x)\ {\frac {dx}{du}}\right)g'(x)\ dx$	(3)	i.e. ${\frac {du}{dx}}=g'(x)$
	$=\int \limits _{x=a}^{x=b}h(g(x))g'(x)dx$	(4)	i.e. Now equate $\left(f(x)\ {\frac {dx}{du}}\right)$ with $h(g(x))$
	$=\int \limits _{x=a}^{x=b}h(u)g'(x)dx$	(5)	i.e. $g(x)=u$
	$=\int \limits _{u=g(a)}^{u=g(b)}h(u)du$	(6)	i.e. $du={\frac {du}{dx}}dx=g'(x)dx$
	$=\int \limits _{u=c}^{u=d}h(u)du$	(7)	i.e. We have achieved our desired result

Example 1

$\int _{0}^{2}x(x^{2}+1)^{2}\operatorname {d} x$

Instead of making this a big polynomial we will just use the substitution method.

Step 1

Identify your u

Let $u=x^{2}+1$

Step 2

Identify $\operatorname {d} u$

$\operatorname {d} u=2x\operatorname {d} x$

Step 3

Now we plug in our limits of integration to our u to find our new limits of integration

When $x=0,u=0^{2}+1=1$

and when $x=2,u=2^{2}+1=5$

Now our integration problem looks something like this

{\frac {1}{2}}\int _{0}^{5}(x^{2}+1)^{2}(2x)\operatorname {d} x

Step 4

write your new integration problem

When we plug in our u it looks like

{\frac {1}{2}}\int _{0}^{5}(u)^{2}\operatorname {d} u

Step 5

Evaluate the Integral

{\frac {1}{2}}\left[{\frac {1}{3}}u^{3}\right]_{0}^{5}

{\frac {1}{2}}\left[\left({\frac {1}{3}}*5^{3}\right)-\left({\frac {1}{3}}*0^{3}\right)\right]

{\frac {1}{2}}\left[{\frac {1}{3}}*125\right]

{\frac {1}{2}}\left[{\frac {125}{3}}\right]

{\frac {125}{6}}

As you can see this all simplified fairly nice. Using substitution will be hard, for most people, at first. Once you get the hang of doing this it should come to you faster and faster each time.

Example 2

\int 3x^{2}(x^{3}+1)^{5}dx

we see that $3x^{2}$ is the derivative of $x^{3}+1$ . Letting

u=x^{3}+1

we have

{\frac {du}{dx}}=3x^{2}

or, in order to apply it to the integral,

du=3x^{2}dx

With this we may write

\int 3x^{2}(x^{3}+1)^{5}dx=\int u^{5}du={\frac {u^{6}}{6}}+C={\frac {(x^{3}+1)^{6}}{6}}+C

Note that it was not necessary that we had exactly the derivative of $u$ in our integrand. It would have been sufficient to have any constant multiple of the derivative.

For instance, to treat the integral