Calculus Of One Real Variable – By Pheng Kim Ving Chapter 10: Techniques Of Integration – Section 10.8: The Method Of Integration By Parts 10.8 The Method Of Integration By Parts

Go To Problems & Solutions

 1. Introduction

yielding:

Wow! … Wait a minute, let's check to see if it's correct: (x sin x + cos x + C)' = (1)sin x + x cos xsin x + 0 =
x cos x. It's correct!

 2. The Method Of Integration By Parts

## Example 2.1

Calculate:

### EOS

#### General Case

Formula [2.1] can be used to find the integral of the product of 2 functions. Suppose that we're given the integral:

The constant of integration for v disappears. As a consequence it isn't necessary. It can only be cumbersome, and hence
should be omitted.

Recall that the method of substitution, a feature of integration, is derived from and inverse to the chain rule, a feature of
differentiation. Now we see that the method of integration by parts, a feature of integration, is derived from and inverse
to the product rule, a feature of differentiation (differentiating uv to get to u'v + uv'; integrating u'v + uv' to get back to
uv).

 3. Applying More Than Once

## Example 3.1

Compute:

### where C = – 2C1. EOS

Here we apply the method of integration by parts twice. First, we apply it to the original integral, and second, to the
integral obtained from the first application. We use the same letters u and v in both steps. Don't be confused. It's ok to do
so. Think of this situation as assigning new values to u and v.

We would have undone what we've done in the first step. This shows that we should make a similar choice for u in the
second application, u = x, as in the first, u = x2.

 4. Regarding The Integrand As The Product Of Itself And 1

## Example 4.1

Evaluate:

### SolutionLet u = lnx and dv = dx, so that du = (1/x) dx and v = x. Thus:

EOS

We have to regard the integrand as the product of itself and 1: ln x = ( ln x) . 1. So we choose u = ln x and dv = 1 . dx
= dx. Looking at it in another way, we can say that sometimes it's necessary to choose dv = dx only.

 5. Re-Appearance Of The Original Integral

## Example 5.1

Find:

Solution
Let u = ex and dv = sin x dx, so that du = ex dx and v = – cos x. Thus:

where C = (1/2)C1.

EOS

 6. Handling The Definite Integrals

## Example 6.1

Calculate:

### Solution EOS

When handling definite integrals, remember to include the evaluation symbol with any term that's been integrated.

 7. Reduction Formulas

## Example 7.1

Derive this formula:

Solution
Let u = lnn x and dv = dx, so that du = (n/x)lnn–1 x dx and v = x. Thus:

EOS

 8. Remarks

The preceding examples show that if the integrand involves a power of x or an inverse trigonometric function or a
logarithm (which is the inverse of the exponential function), then we should let u be the power of x (to lower the power)
or the inverse trigonometric function or the logarithm (to transform the integrand into an algebraic expression). See
Example 3.1 for a power of x, Example 6.1 for an inverse trigonometric function, and Example 4.1 for a logarithm.

##### Problems & Solutions

1. Calculate the following integrals.

Solution

a.  Let u = x and dv = sin 5x dx, so that du = dx and v = –(1/5) cos 5x. Thus:

where C = – C1.

It follows that:

Let u = e2x and dv = cos 3x dx, so that du = 2e2x dx and v = (1/3) sin 3x. Thus:

where C = (9/13)C1.

2. Compute this definite integral:

Solution

3. Find a reduction formula for:

Solution

Let u = xn and dv = eax dx, so that du = nxn–1 dx and v = (1/a) eax. Thus:

4. Let:

Solution

a. Let u = sinn–1 x and dv = sin x dx, so that du = (n – 1)sinn–2 x cos x dx and v = – cos x. Then:

where C = (3/4)C1,

where C = (5/6)C1.

5. Let:

Solution

c. We have: