4.5Integration by Substitution

Antidifferentiation of a Composite Function

Let f and g be functions such that f og and g’ arecontinuous on an interval I. If F is an antiderivativeof f on I, then

∫ += CxgFdxxgxgf ))(()('))((

Outside

Inside

Derivative of Inside

Evaluate

€

x 2 +1( )2

2x( )dx∫ Let u = x2 + 1

du = 2x dx

dxx

du=

2

€

u( )2

2x( )du

2x∫

Cu

+=3

3

€

=x 2 +1( )

3

3+C

Evaluate dxx5cos5∫ Let u = 5xdu = 5 dx

dxdu

=55

cos5ud

u∫=Cu +=sin Cx += 5sin

Multiplying and dividing by a constant

€

x x 2 +1( )∫2dx Let u = x2 + 1

du = 2x dx

dxx

du=

2( )x

duux

2

2

∫= Cu

+=6

3

€

=x 2 +1( )

3

6+C

dxx∫ −12 Let u = 2x - 1

du = 2dx

dxdu

=2

2

duu∫= 2

21 duu∫=

Cu

+=3

221 23

€

=2x −1( )

3 2

3+C

Substitution and the General Power RuleWhat would you let u = in the following examples?

( )( )

( )∫

∫

∫∫∫

−−

−

++

−

dxxx

dxx

x

dxxx

dxxxx

dxx

sincos

21

4

23

12

)13(3

2

2

32

2

4 u = 3x - 1

u = x2 + x

u = x3 - 2

u = 1 – 2x2

u = cos x

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

A differential equation, a point, and slope field are given. Sketchthe solution of the equation that passes through the given point.Use integration to find the particular solution of the differential equation.

€

dy

dx= x 2(x 3 −1)2 1,0( )

€

x 2 x 3 −1( )2

∫ dx

€

u = x 3 −1

du = 3x 2dx

du

3x 2= dx

€

x 2 u( )2∫ du

3x 2

€

u2du

3∫ =

u3

9+C =

x 3 −1( )3

9+C

€

0 =13 −1

9+C

C = 0

€

y =x 3 −1( )

3

9 Day 1 stop (1-41 odd)

∫ dxxx 32 sin u = x3

du = 3x2 dx

dxx

du=23

∫=2

2

3sin

x

duux

∫= duusin3

1Cu +−= )cos(

31

Cx +−= 3cos31

∫ dxxx 3cos3sin 2

rewritten as

€

sin 3x( )2cos3x dx∫

Let u = sin 3x

du = 3cos 3x dx

dxx

du=

3cos3

x

duxu

3cos3

3cos2∫= duu∫= 2

31

Cu

+=9

3

Cx+=

93sin3

Day 2 stop (43-56 all, 57-61 odd)

dxxx∫ −12 Let u = 2x - 1

du = 2dx

dxdu

=2

u + 1 = 2x

xu

=+21

22

1 duuu∫ ⎟

⎠

⎞⎜⎝

⎛ +=

€

=1

4u3 2 + u1 2

( )∫ du Cuu

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

32

52

41 2325

( ) ( )C

xx+

−+

−=

6

12

10

12 2325

Evaluate

dxxx∫ +1

0

32 )1(

u = x2 + 1

du = 2x dx dxx

du=

2

x

duux

2)( 3∫=

Note that there are no upper and lowerlimits of integration.

⎥⎦

⎤=

8

4u

We must determine new upper and lower limits by substituting the oldones in for x in u = x2 + 12

1

8

15

8

1

8

16=−=

Or, we could use the old limits if we substitute x2 + 1 back in.

€

=x 2 +1( )

4

8

⎤

⎦

⎥ ⎥0

1

8

15

8

1

8

16=−=

dxx

xArea ∫ −

=5

1 12

( )uduu

u∫

+=

2

12

12 −= xu

122 −= xu2u du = 2 dx

u du = dx2

12 +=

ux

( )duu∫ += 12

1 2⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛+= u

u

32

1 3

What limits arewe going to use?3

1

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ +−⎟⎠

⎞⎜⎝

⎛ += 13

13

3

27

2

13

16=

See area comparisons when using different upper and lower limits. Page 302

Integration of Even and Odd Functions

If f is an even function, then

∫ ∫−

=a

a

a

dxxfdxxf0

)(2)(

If f is an odd function, then

∫−

=a

a

dxxf 0)(

Ex. ∫−

2

2

3dxx

Odd or Even?

0=