Integration by Substitution[edit | edit source]
There is a theorem that will help you with substitution for integration. It is called Change of Variables for Definite Integrals.
what the theorem looks like is this
In order to get
you must plug a into the function g and to get
you must plug b into the function g.
The tricky part is trying to identify what you want to make your u to be. Some times substitution will not be enough and you will have to use the rules for integration by parts. That will be covered in a different section
Ex. 1

Instead of making this a big polynomial we will just use the substitution method.
Step 1
Identify your u
Let 
Step 2
Identify 

Step 3
Now we plug in our limits of integration to our u to find our new limits of integration
When 
and when 
Now our integration problem looks something like this
Step 4
write your new integration problem
When we plug in our u it looks like
Step 5
Evaluate the Integral
![{\displaystyle {\frac {1}{2}}\left[\left({\frac {1}{3}}*5^{3}\right)-\left({\frac {1}{3}}*0^{3}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d2d3071a368bc21723f39d733a5bea9707fee02)
![{\displaystyle {\frac {1}{2}}\left[{\frac {1}{3}}*125\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e88b46a59ca7f729f5d4fe802a9e4d1ccce9c928)
![{\displaystyle {\frac {1}{2}}\left[{\frac {125}{3}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5d5e2d3f3ed81959a9a43ee6fc689e137285a62)

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.
I'll give you some other problems to work on as well.
Ex. 2

Ex. 3

Ex. 2
Let 
Then 
When x = 0

and when

Therefore,

![{\displaystyle \left[{\frac {1}{2}}u^{2}\right]_{0}^{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/778dff182da7fd7e83412172a77ab3133290439f)
![{\displaystyle \left[{\frac {1}{2}}\sin ^{2}(1)\right]-\left[{\frac {1}{2}}\sin ^{2}(0)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faca3d73949ce418c705831f25d37438a3ca186e)

Ex. 3
Let 
Then 
plug in our limits to get new limits
When x = -1

and when x = 2

Our new integration problem is

Giving us
![{\displaystyle =\left[{\frac {2}{3}}*(x^{2}+4)\right]_{5}^{8}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94dd69ed5a7f9521043490005cc8954f41854b9f)
![{\displaystyle =\left[{\frac {2}{3}}*(8^{2}+4)\right]-\left[{\frac {2}{3}}*(5^{2}+4)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e62770a99b361040bf607abca6950f15e07024f4)
![{\displaystyle =\left[{\frac {136}{3}}\right]-\left[{\frac {58}{3}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f8f4f1c858313a7e9f1eec53d5f245d5446483c)
