﻿ ﻿STANDARD ELEMENTARY INTEGRAL FORMS - General Methods of Integration - The Calculus Primer

## The Calculus Primer (2011)

### Chapter 46. STANDARD ELEMENTARY INTEGRAL FORMS

12—8. Standard Integrals. Thus far we have discussed five so-called standard elementary integrals, which we restate below for convenience: dx = x + C a dv = a dv (du + dv − dw) = du + dv − dw We now give a list of the remaining standard integral forms for reference:   12—9. The Forms avdv and ev dv. These formulas are seen to be true from the following considerations. We know that

d(av) = log a·av·dv;

integrating both sides, we get:

av = log a·av·dv,

or av = log a av dv;

hence In a similar manner, since

d(ev) =evdv,

it follows that ev dv = ev + C.

EXAMPLE 1. Find e2x dx.

Solution. If we let v = 2x, then dv = 2dx. Now, in order to make the expression e2x dx conform to the standard form ev dv, since dv = 2dx when v = 2x, we insert the factor 2 before the dx, and the factor before the integral sign; thus but ev dv = ev + C, hence EXAMPLE 2. Find 2a3x dx.

Solution. Let v = 3x; then dv = 3dx. Insert 3 before the dx, and before the integral sign; then EXAMPLE 3. Find ecos x (sin x) dx.

Solution. Let v = cos x; dv = − sin x dx; then ecos x (sin x dx) = ecos x (− sin x dx) = − ecos x + C.

EXERCISE 12—4

Find the following integrals; check by differentiation:

1.  e4z dx

2. kamx dx

3. esin x (cox x) dx

4. e2 sin x (cos x) dx

5. e−3x dx

6. ax/n dx

7. a2x−1 dx

8. e2x3·x2 dx

9. ex (ex + 1)dx

10.  11. xkx2 dx

12. (ex/2 + ex/2)dx

12—10. Standard Forms  through . These formulas follow immediately from the corresponding formulas for differentiation, as given in §5—9, §5—10, §5—11, §5—12, equations –. They may be verified simply and directly by differentiation.

EXAMPLE 1. Find cos 3ax dx.

Solution. Let v = 3ax; then dv = 3a dx. EXAMPLE 3. Find csc2x2·x dx.

Solution. Let v = x2; then dv = 2x dx.  EXERCISE 12—5

Find the following integrals; check by differentiation:  Verify the following: 12—11. Standard Forms –. These four formulas may be proved by transforming the integrand in each case so that we may apply equation ; the proofs follow.

Form : But − log cos v = − = − log 1 + log sec v = log sec v;

hence, tan v dv = log sec v.

Form : In a similar manner, Form : To transform the integrand so that it will be in the form we write:  Form : This may be derived in a manner similar to the proof for . We leave it as an exercise for the reader; multiply the integrand csc v by the fraction .

12—12. Forms –. The formulas for the standard forms – may be derived by suitable transformations or substitutions. Formulas  and  follow at once from the corresponding formulas for differentiation.

EXAMPLE 1. Find Solution. Let v2 = 9x2, and a2 = 4; then v = 3x, dv = 3dx, and a = 2.

From formula , we get: EXAMPLE 2. Find Solution. Let v2 = 4x2, and a2 = 25; then v = 2x, dv = 2dx, and a = 5.

From formula , we get: EXAMPLE 3. Find Solution. Let v2 = 4x2, and a2 = 9; then v = 2x, dv = d(2x), and a = 3.

From formula : EXAMPLE 4. Find .

Solution. Let v2 = 5x2, and a2 = 3; then v = x, dv = dx, and a = .

From formula : EXAMPLE 5. Find Solution. Rewrite as follows: Here v = 2x, a = 5, dv = 2dx.

From formula : EXAMPLE 6. Find Solution. Let v2 = x4, and a2 = 36; then v = x2, dv = 2x dx, and a = 6.

From formula : EXAMPLE 7. Find Solution. Rewrite as follows, completing the square in the denominator: Now apply formula , where v = x + 2, a = 4. EXERCISE 12—6

Find the following integrals; check by differentiation:  Verify the following:  EXERCISE 12—7

Review

Find the following integrals:  13. ecos x.sin x dx

14. The slope of a certain curve is given by = 6x2 − 10x + 8. If the curve passes through the point (2,10), find its equation.

15. If = 12x + 6, find y in terms of x if it is known that when x = 2, = 28, and when x = −3, y = − 1.

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