Section 8.2 – Integration by Parts. Find the Error The following is an example of a student response. How can you tell the final answer is incorrect?

Section 8.2 – Integration by Parts

Find the ErrorThe following is an example of a student response. How can you tell the final answer is incorrect? Where did the student make an error?

Evaluate:

This is not the antiderivative of since

2

cos cos

1sin

2

x x dx x dx x dx

x x C

The integral of a product is not equal to product of the integrals.

This should remind us of the Product Rule. Is there a way to use the Product Rule to investigate the

antiderivative of a product?

Integration by Parts: An Explanation

When u and v are differentiable functions of x:

d dv dudx dx dxuv u v The Product

Rule tells us…

d dv dudx dx dxuv dx u dx v dx If we integrate

both sides…

uv u dv v du If we simplify the integrals…

u dv uv v du If we solve for one of the integrals…

Integration by Parts

When evaluating the integral and f (x)dx=u dv (with u and v being differentiable functions of x), then, the following holds:

u dv uv v du Rewrite the function into

the product of u and dv.

The integral equals…

u times the antiderivative

of dv.

The integral of the product of

the antiderivative of dv and the

derivative of u.

Integration by Parts: The Process

Since f (x)dx=u dv, success in using this important technique depends on being able to separate a given integral into parts u and dv so that…

a) dv can be integrated.

b) is no more difficult to calculate than the original integral.

The following does NOT always hold, but is very helpful:

Frequently, the derivative of u, or any higher order derivative, will be zero.

Example 1

Evaluate:

u Pick the u and dv.

dv x cos x dx

du Find du and v. v dx sin x

cosx x dx Apply the formula. sinx x sin x dx

sin cosx x x C

Differentiate. Integrate.

Example 2

Evaluate: u Pick the u

and dv.dv 2x xe dx

du Find du and v. v 2xdx xe

2 xx e dx Apply the formula.

2 xx e 2 xxe dx Differentiate. Integrate.

You may need to apply

Integration by Parts Again.

u Pick the u and dv.

dv 2x xe dx

du Find du and v. v 2dx xe

2 xx e Apply the formula. 2 xxe 2 xe dx

2 2 2x x xx e xe e C

2 xxe dx

White Board Challenge

Evaluate:

3 2323 1x x x 5 224

15 3 1 1x x 7 21635 1x x 9 232

315 1x C

Example 3

Evaluate: Since, multiple Integration by Parts are needed, a Tabular Method is a

convenient method for organizing repeated Integration by parts.

Repeated Differentiation Repeated Integration3x x23 1x 6x

60

1 21x 3 22

3 1x 5 24

15 1x 7 28

105 1x 9 216

945 1x

+–+

–

3 2323 1x x x 5 224

15 3 1 1x x 7 21635 1x x 9 232

315 1x C

Must get 0.

Start with +Alternate

Find the sum of the products of each diagonal:

Differentiate the u.

Integrate the dv.

Connect the

diagonals.

Notice the cubic function will go to zero. So it is a good choice for u.

Example 4

Evaluate:

u Pick the u and dv.

dv x ln x dx

du Find du and v. v dx ??

This was a bad choice for u and dv.

Differentiate. Integrate.

Example 4: Second Try

Evaluate:

u Pick the u and dv.

dv ln x x dx

du Find du and v. v 1

x dx 212 x

lnx x dx Apply the formula.

212 lnx x 1

2 x dx 2 21 1

2 4lnx x x C

Differentiate. Integrate.

Try the opposite this time.

Example 5

Evaluate:

1 ln x dxIf there is only one function, rewrite the integral so there is two.

u Pick the u and dv.

dv ln x 1dx

du Find du and v. v 1

x dx x

ln x dx Apply the formula. lnx x 1dx

lnx x x C

Differentiate. Integrate.

Example 6

Evaluate: u Pick the u and dv. dv cos x xe dx

du Find du and v. v sin x dx xecosxe x dx Apply the

formula. cosxe x sinxe x dx You may need

to apply Integration by Parts Again.

cos cosx xe x dx e x Apply the formula. sinxe x cosxe x dx

1 12 2cos cos sinx x xe x dx e x e x C

cos cos sinx x xe x dx e x e x dx sinxe x dxu Pick the u and dv. dv sin x xe dx

du Find du and v. v cos x dx xe

2 cos cos sinx x xe x dx e x e x If you see the integral you are

trying to find, solve for it.

Example 7

Evaluate: 1 1

01 tan x dxIf there is only one function, rewrite the

integral so there is two.

u Pick the u and dv.

dv 1tan x 1dx

du Find du and v. v 2

11 x

dx

x1 1

01 tan x dx Apply the

formula. 11

0tanx x

2

1

10

xx

dx

1

214 2 0

ln 1 x 1

4 2 ln 2

Integration by Parts: Helpful Acronym

When deciding which product to make u, choose the function whose category occurs earlier in the list below. Then take dv to be the rest of the integrand.

L

I

A

T

E

ogarithmicnverse trigonometric

lgebraic

rigonometric

xponential

White Board Challenge

Evaluate:

3 2432 1 1x x x C