﻿ ﻿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 [1]

a dv = a dv [2]

(du + dv − dw) = du + dv − dw [3]

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 [8] through [13]. These formulas follow immediately from the corresponding formulas for differentiation, as given in §5—9, §5—10, §5—11, §5—12, equations [6]–[11]. 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 [14]–[17]. These four formulas may be proved by transforming the integrand in each case so that we may apply equation [5]; the proofs follow.

Form [14]:

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

hence, tan v dv = log sec v.

Form [15]: In a similar manner,

Form [16]: To transform the integrand so that it will be in the form we write:

Form [17]: This may be derived in a manner similar to the proof for [16]. We leave it as an exercise for the reader; multiply the integrand csc v by the fraction .

12—12. Forms [18]–[23]. The formulas for the standard forms [18]–[23] may be derived by suitable transformations or substitutions. Formulas [22] and [23] 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 [18], we get:

EXAMPLE 2. Find

Solution. Let v2 = 4x2, and a2 = 25; then v = 2x, dv = 2dx, and a = 5.

From formula [19], we get:

EXAMPLE 3. Find

Solution. Let v2 = 4x2, and a2 = 9; then v = 2x, dv = d(2x), and a = 3.

From formula [20]:

EXAMPLE 4. Find .

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

From formula [19]:

EXAMPLE 5. Find

Solution. Rewrite as follows:

Here v = 2x, a = 5, dv = 2dx.

From formula [22]:

EXAMPLE 6. Find

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

From formula [20]:

EXAMPLE 7. Find

Solution. Rewrite as follows, completing the square in the denominator:

Now apply formula [18], 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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