The Calculus Primer (2011)
Part XII. General Methods of Integration
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.
The reader can readily verify these results by differentiation.
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 + e−x/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.