Integration by Substitution

INTEGRATION BY SUBSTITUTION

Section 6.2a

A change of variables can often turn anunfamiliar integral into one that we canevaluate…

This method is called thesubstitution method of integration.

The New Method

The New MethodGenerally, this method is used when integrating a compositefunction, and the derivative of the inside function is alsopresent in the integrand:

f g x g x dx f u du

1. Substitute u = g(x), du = g (x)dx

F u C 2. Evaluate by finding an antiderivative F (u) of f (u)

F g x C 3. Replace u by g(x)

Initial Practice ProblemsEvaluate 52x dxLet 2u x Then 2du d x dx

Substitute: 5 52x dx u du 6

6u C

626

xC

(Substitute)

Initial Practice ProblemsEvaluate 4 1x dx

Let 4 1u x Then 4du dx14

dx du

Substitute: 1 2 14 14

x dx u du

1 214u du 3 21 2

4 3u C

3 216u C 3 21 4 1

6x C

Initial Practice ProblemsEvaluate cos 7 5x dx

Let 7 5u x Then 7du dx17

dx du

Substitute: 1cos 7 5 cos7

x dx u du

1 cos7

u du 1 sin7

u C

1 sin 7 57

x C

Initial Practice ProblemsEvaluate tan x dxLet cosu x Then sindu xdx

Substitute:sincosx dudxx u

duu

ln u C

sincosx dxx

1ln Cu

1lncos

Cx

ln sec x C

Substitution in Definite IntegralsInstead of the last step we’ve been using (re-substitution???):

Substitute , and integrate with u g x du g x dxrespect to u from to . u g a u g b

b g b

a g af g x g x dx f u du

Evaluating Definite IntegralsEvaluate

4 2

0tan secx xdx

Let tanu x Then 2secdu xdx

Also, notice: 0 tan 0 0u tan 14 4

u

Substitute:4 12

0 0tan secx xdx udu

12

02u

1 102 2

Two Possible Methods?Evaluate

1 2 3

13 1x x dx

Let 3 1u x Then 23du x dx

Also, notice: 1 0u 1 2u

23 2

0

23u

4 23

Method 1

Substitute: 1 22 3

1 03 1x x dx udu

Two Possible Methods?Evaluate

1 2 3

13 1x x dx

Let 3 1u x Then 23du x dx

13 2

1

23

x

x

u

13 23

1

2 13x

Substitute:1 12 3

1 13 1

x

xx x dx udu

4 23

Method 2

Guided PracticeEvaluate the given integral.

2

3

9

1

r dr

r

31u r 36 1 r C 23du r dr

213du r dr

1 2193

u du

1 23 2u C

Guided PracticeEvaluate the given integral.

sinu xcosdu xdx

4u du51

5u C

4sin cosx x dx

5sin5x C

Guided PracticeEvaluate the given integral.

7

0 2dxx 2u x

9ln 1.5042

du dx

0 2u

7 9u

9

2

1 duu

9

2ln u

ln 9 ln 2

Guided PracticeEvaluate the given integral.

tanu x2secdu xdx

14

u

14

u

1 2

1u du

4 2 2

4tan secx xdx

23

13

13u

1 13 3

Guided PracticeEvaluate the given integral.

lnu x1du dxx

1u e

6 ln 6u

ln6

1

1 duu

ln 6

1ln u

ln ln 6 ln1

6

lne

dxx x

ln ln 6 0.583