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

The Calculus Primer (2011)

Part XIII. Special Methods of Integration

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 image

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

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

By integrating both sides of (2), we obtain

image u dv = uv − image v du. [1]

It will be seen that we may make use of equation [1] 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:

image f(x) dx = image u dv; image u dv = uv − image v du. [2]

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 image v du is easier to evaluate than the original integral, image 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 image x sin x dx.

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

Substituting in equation [2] above:

image x sin x dx = x cos x − image cos x dx

= − x cos x + sin x + C.

EXAMPLE 2. Find image x log x dx.

Solution. Let u = log x, and dv = x dx; then image, and image

Substituting in [2]:

EXAMPLE 3. Find image xex dx.

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

Hence: image xex dx = xeximage ex dx

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

EXAMPLE 4. Find image arc sin x dx.

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

Hence, image arc sin x dx = x arc sin ximage image

But, by §12—6, Example 2, we see that

hence image arc sin x dx = x arc sin image

EXAMPLE 5. Find image x2 sin x dx.

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

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

But, to find the integral image 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, image x cos x dx = x sin ximage sin x dx

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

Therefore, substituting (2) in (1):

image 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 image x2 cos 2x dx.

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

But, to find the integral image 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. image x cos x dx

2. image log x dx

3. image xe2z dx

4. image x2exdx

5. image ex cos x dx

6. image log2 x dx

7. image x2 cos x dx

8. image x3 log x dx

9. image θ sec2 θ dθ

10. image arc tan x dx

11. image x2 e−x dx

12. image x2 log x dx

13. image θ tan2 θ dθ

14. image z sec2 z dz

15. image ex cos 2x dx

16. image cos x log sin x dx

17. image x2 sin 2x dx

18. image x cos 2x dx