﻿ ﻿INTEGRATION BY PARTS - Special Methods of Integration - The Calculus Primer

## The Calculus Primer (2011)

### Chapter 47. INTEGRATION BY PARTS

13—1. Need for Special Methods. The standard forms discussed in the preceding chapter enable us to integrate many expressions that arise. However, other types of expressions may be encountered for which these standard forms will not suffice, and so certain special methods must be employed. One of the most common of these is the method of integration by parts.

13—2. Integration by Parts. Consider u and v, which represent functions of x. We have: which may also be written in the differential form or d(uv) = v du + u dv, (1)

where Now, by transposing, equation (1) may be written

u dv = d(uv) v du. (2)

By integrating both sides of (2), we obtain u dv = uv − v du. 

It will be seen that we may make use of equation  whenever it is possible to find the integral of v du. The method of “integrating by parts” thus amounts to this: if we cannot integrate f(x) dx directly, we try to break the expression f(x) dx into two factors or “parts,” say u and dv, such that the integrals of both dv and v du are readily found. The method of procedure is suggested by the following: f(x) dx = u dv; u dv = uv − v du. 

Frankly, no general rule can be given to indicate the best way of selecting the factors u and dv in all cases; each example must be studied individually. However, with practice the reader will doubtless develop facility in using this method, which is one of the most useful of all special methods of integration. It may help to remember that we try to select u, and the corresponding dv, in such a way that not only can we find v from dv, but also that v du is easier to evaluate than the original integral, f(x) dx. In some cases it may be necessary to repeat the process one or more times. The following examples will illustrate the application of the method of integration by parts.

EXAMPLE 1. Find x sin x dx.

Solution. Let u = x, and dv = sin x dx; then du = dx, and v = sin x dx = − cos x.

Substituting in equation  above: x sin x dx = x cos x − cos x dx

= − x cos x + sin x + C.

EXAMPLE 2. Find x log x dx.

Solution. Let u = log x, and dv = x dx; then , and Substituting in : EXAMPLE 3. Find xex dx.

Solution. Let u = x, and dv = ex dx; then du = dx, and v = ex.

Hence: xex dx = xex ex dx

= xexex + C = ex(x1) + C.

EXAMPLE 4. Find arc sin x dx.

Solution. Let u = arc sin x, and dv = dx; then and v = x.

Hence, arc sin x dx = x arc sin x  But, by §12—6, Example 2, we see that hence arc sin x dx = x arc sin EXAMPLE 5. Find x2 sin x dx.

Solution. Let u = x2, and dv = sin x dx; then du = 2x dx, and v= sin x dx = − cos x.

Hence: x2 sin x dx = −x2 cos x + 2 x cos x dx. (1)

But, to find the integral x cos x dx in equation (1), we must apply the method of integration by parts again; thus

let u = x, and dv = cos x dx;

du = dx, and v = sin x.

Hence, x cos x dx = x sin x sin x dx

= x sin x + cos x + C. (2)

Therefore, substituting (2) in (1): x2 sin x dx = −x2 cos x + 2(x sin x + 2 cos x) + C

= −x2 cos x + 2x sin x + 2 cos x + C′.

EXAMPLE 6. Find x2 cos 2x dx.

Solution. Let u = x2, and dv = cos 2x dx; then du = 2x dx, and  But, to find the integral x sin 2x dx, we use the method once more:

let u = x, and dv = sin 2x dx;  Therefore, substituting (2) in (1): EXERCISE 13—1

Find the following integrals, using the method of integration by parts:

1. x cos x dx

2. log x dx

3. xe2z dx

4. x2exdx

5. ex cos x dx

6. log2 x dx

7. x2 cos x dx

8. x3 log x dx

9. θ sec2 θ dθ

10. arc tan x dx

11. x2 e−x dx

12. x2 log x dx

13. θ tan2 θ dθ

14. z sec2 z dz

15. ex cos 2x dx

16. cos x log sin x dx

17. x2 sin 2x dx

18. x cos 2x dx

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