Integration byIntegration bySubstitutionSubstitution

The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions? We can sometimes use substitution or change of variable to rewrite functions in a form that we can integrate.

Example 1:

52x dx Let 2u x

du dx

The variable of integration must match the variable in the expression.

Don’t forget to substitute the value for u back into the problem!

duu5

cu 6

6

1

cx

6

)6( 6

Example 2:

21 2 x x dx One of the clues that we look for is if we can find a function and its derivative in the integral.

The derivative of is .21 x 2 x dx

1

2 u du3

22

3u C

3

2 22

13

x C

2Let 1u x

2 du x dx

Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.

Example 3:

4 1 x dx Let 4 1u x

4 du dx

1

4du dx

Solve for dx.1

21

4

u du3

22 1

3 4u C

3

21

6u C

3

21

4 16

x C

Example 4:

cos 7 5 x dx7 du dx

1

7du dx

1cos

7u du

1sin

7u C

1sin 7 5

7x C

Let 7 5u x

Example 5:

2 3sin x x dx 3Let u x23 du x dx

21

3du x dx

We solve for because we can find it in the integral.

2 x dx

1sin

3u du

1cos

3u C

31cos

3x C

Example 6:

4sin cos x x dx

Let sinu x

cos du x dx

4sin cos x x dx

4 u du51

5u C

51sin

5x C

Example 7:

24

0tan sec x x dx

The technique is a little different for definite integrals.

Let tanu x2sec du x dx

0 tan 0 0u

tan 14 4

u

1

0 u du

We can find new limits, and then we don’t have to substitute back.

new limit

new limit

We could have substituted back and used the original limits.

1

0

2

2

1

u

2

1

Example 7 continued:

24

0tan sec x x dx

Let tanu x

2sec du x dx4

0 u du

Wrong!The limits don’t match!

221 1

tan tan 02 4 2

2 21 11 0

2 2

u du21

2u

1

2

Using the original limits:

Leave the limits out until you substitute back.

This is usually more work than finding new limits

4

0

2)(tan2

1

x

Example 8:

1 2 3

13 x 1 x dx

3Let 1u x

23 du x dx 1 0u

1 2u 1

22

0 u du

Don’t forget to use the new limits.

22 2

3 4 2

3

2

0

2

3

3

2

u

2

3

23

2

Acknowledgement

I wish to thank Greg Kelly from Hanford High School, Richland, USA for his hard work in creating this PowerPoint.

http://online.math.uh.edu/

Greg has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au