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39.8 Compute the value of
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sin nx cos kx dx for positive integers n and k.
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n = k.
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By Problem 39.7, with A = n and B = k,
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sin nx cos kx dx =
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1 2 (sin (n + k)x + sin (n k)x) dx 2 0 cos (n k)x 2 1 cos (n + k)x + =0 = 2 n+k n k 0
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because cos px is, for p an integer, a periodic function of period 2 . Case 2: n = k. Then, by the double-angle formula for the sine function,
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sin nx cos nx dx =
1 2 1 sin 2nx dx = 2 0 2
cos 2nx 2n
39.9 Find
dx x2 + 9
x 2 + 9 suggests letting x = 3 tan . Then dx = 3 sec2 d , and x2 + 9 = 9 tan2 + 9 = 9(tan2 + 1) = 3 sec2 = 3 sec
The presence of
CHAP. 39]
TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
dx x2 + 9
= = ln = ln
3 sec2 d = 3 sec
sec d = ln |sec + tan | + C x2 + 9 + x +C 3
x x2 + 9 + + C = ln 3 3
x 2 + 9 + x + K = ln( x 2 + 9 + x) + K
note
x2 + 9 + x = ln 3
x 2 + 9 + x ln 3
and the constant ln 3 can be absorbed in the arbitary constant K. Furthermore, x2 + 9 + x > 0
39.10 Find
dx . 3 x2
3 x 2 suggests the substitution x = 3 x2 = 3 3 sin2 =
The presence of
3 sin . Then, dx = 3 cos d . 3 cos2 = 3 cos
3(1 sin2 ) =
and x2
dx 3 x2
1 1 3 cos d d = = 3 3 (3 sin2 )( 3 cos ) sin2 1 = cot + C 3 cos 3 x2 / 3 3 x2 = = cot = sin x x/ 3 dx x2 3 x2 = 3 x2 +C 3x
csc2 d
Hence,
39.11 Find
x2 dx. x2 4
x 2 4 suggests the substitution x = 2 sec . Then dx = 2 sec tan d . x2 4 = 4 sec2 4 = 4(sec2 1) = 2 tan2 = 2 tan sec3 d [by Problem 38.13(b)]
The occurrence of
x2 x2 4
dx =
(4 sec2 )(2 sec tan ) d = 4 2 tan x2 4 2 x2 4 2
= 2(sec tan + ln |sec + tan |) + C =2 = = x 2 + ln x + 2
x x2 4 x + x2 4 +C + 2 ln 2 2 x x2 4 + 2 ln x + 2 x2 4 + K
where K = C 2 ln 2 (compare Problem 39.9).
TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
[CHAP. 39
Supplementary Problems
39.12 Find the following antiderivatives: (a) (e) (i) (m) sin x cos2 x dx cos6 x sin2 x dx tan2 x sec4 x dx sin x cos 3 x dx (b) (f ) (j) (n) cos2 3x dx tan2 x dx 2 (c) (g) (k) (o) sin4 x cos5 x dx tan6 x dx tan4 x sec x dx cos 4x cos 9x dx (d) (h) (l) cos6 x dx sec5 x dx sin 2x cos 2x dx
tan3 x sec3 x dx sin 5x sin 7x dx
39.13 Prove the following identities: 1 (a) sin Ax sin Bx = (cos (A B)x cos (A + B)x) 2 1 (b) cos Ax cos Bx = (cos (A B)x + cos (A + B)x) 2 39.14 Calculate the following de nite integrals, where the positive integers n and k are distinct:
2 2
sin nx sin kx dx
sin2 nx dx
39.15 Evaluate: (a) (e) (i) x2 1 dx x dx x2 x2 9 (b) (f ) (j) x2 4 x2 dx (c) (g) (k) 1 + x2 dx x dx (x 2 + 9)2 dx (x 2 6x + 13)2 (d) (h) x 2 x2 dx dx
dx (4 x 2 )3/2 e3x 1 e2x dx
x 16 9x 2
x 2 1 x 2 dx
[Hint: In part (k), complete the square.] 39.16 Find the arc length of the parabola y = x 2 from (0, 0) to (2, 4). 39.17 Find the arc length of the curve y = ln x from (1, 0) to (e, 1). 39.18 Find the arc length of the curve y = ex from (0, 1) to (1, e). 39.19 Find the are length of the curve y = ln cos x from (0, 0) to ( /3, ln 2). 39.20 Find the area enclosed by the ellipse x2 y2 + = 1. 9 4
Integration of Rational Functions; The Method of Partial Fractions
This chapter will give a general method for evaluating inde nite integrals of the type N(x) dx D(x) where N(x) and D(x) are polynomials. That is to say, we shall show how to nd the antiderivative of any rational function f (x) = N(x)/D(x) (see Section 9.3). Two assumptions will be made, neither of which is really restrictive: (i) the leading coef cient (the coef cient of the highest power of x) in D(x) is +1; (ii) N(x) is of lower degree than D(x) [that is, f (x) is a proper rational function]. EXAMPLES
(a) 8x 4 1 x 10 + 3x 11 7 = 7 56x 4 8x 4 = 10 1 x 10 + 3x 11 7 x 21x + 77 7 x 4 + 7x . Long division (see Fig. 40-1) yields x2 1 7x + 1 f (x) = x 2 + 1 + 2 x 1 Consequently, f (x)dx = (x 2 + 1) dx + x3 7x + 1 dx = +x+ 3 x2 1 7x + 1 dx x2 1
(b) Consider the improper rational function f (x) =
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