sin2 d x2 4 2 x x +C in VS .NET

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sin2 d x2 4 2 x x +C
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The general rule illustrated by this example is: If x = a sec , with 0 < ( /2) or < 3 /2.
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x 2 a2 occurs in an integrand, try the substitution
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39.1 Find sin3 x cos2 x dx.
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The exponent of sin x is odd. So, let u = cos x. Then, du = sin x dx, and sin3 x cos2 x dx = = = = sin2 x cos2 x sin x dx (1 cos2 x) cos2 x sin x dx (1 u2 )u2 du = (u4 u2 ) du
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u2 cos5 x cos3 x u2 +C = +C 5 3 5 3
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TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
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[CHAP. 39
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39.2 Find
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cos4 x sin4 x dx.
The exponents are both even; in addition, they are equal. This allows an improvement on the method of Section 39.1, example (b). sin 2x 4 1 dx = 2 16 1 16 1 64 1 64 1 64 1 64 1 64 1 128 1 cos 4x 2 dx 2 (1 2 cos 4x + cos2 4x) dx 1 dx x 1 2 cos u du + 1 4 cos2 u du [letting u = 4x]
cos4 x sin4 x dx = = = = = = = =
sin4 2x dx
1 1 sin u + (u + sin u cos u) + C 2 8 1 x sin 4x cos 4x x sin 4x + + +C 2 2 8 3x sin 4x sin 8x + +C 2 2 16 sin 8x 3x sin 4x + +C 8
39.3 Find: (a)
cos5 x dx; (b)
sin4 x dx.
(cos2 x)2 (cos x) dx (1 2 sin2 x + sin4 x)(cos x) dx
cos5 x dx = =
cos4 x(cos x) dx =
(1 sin2 x)2 (cos x) dx =
Let u = sin x. Then du = cos x dx, and u5 2u3 + +C 3 5
cos5 x dx =
(1 2u2 + u4 ) du = u
= sin x
2 sin3 x sin5 x + +C 3 5
(b) This antiderivative was essentially obtained in Problem 39.2, x 2
sin4 x dx = 2 =2
sin4 (2u) du
let u =
16 sin 8u 3u sin 4u + +C 128 8 1 3x sin 4x = sin 2x + +C 4 2 8
39.4 Find
tan5 x dx.
CHAP. 39]
TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
From the reduction formula (39.1), tan3 x dx = tan5 x dx = tan2 x 2 tan4 x 4 tan x dx = tan3 x dx = tan2 x ln |sec x| 2 tan2 x tan4 x + ln |sec x| + C 4 2
39.5 Show how to nd tanp x secq x dx: (a) when q is even; (b) when p is odd. (c) Illustrate both techniques with tan3 x sec4 x dx and show that the two answers are equivalent.
(a) Let q = 2r (r = 1, 2, 3, . . .). Then tanp x sec2r x dx = = tanp x sec2(r 1) x(sec2 x) dx tanp x(1 + tan2 x)r 1 (sec2 x) dx
since 1 + tan2 x = sec2 x. Now the substitution u = tan x, du = sec2 x dx, produces a polynomial integrand. (b) Let p = 2s + 1 (s = 0, 1, 2, . . .). Then, tan2s+1 x secq x dx = = tan2s x secq 1 x(sec x tan x) dx (sec2 x 1)s secq 1 x(sec x tan x) dx
since tan2 x = sec2 x 1. Now let v = sec x, dv = sec x tan x dx, to obtain a polynomial integrand. (c) By part (a), tan3 x sec4 x dx = = = By part (b), tan3 x sec4 x dx = = = Since 1 + u2 = v 2 , v6 v4 4v 6 6v 4 4(1 + u2 )3 6(1 + u2 )2 = = 6 4 24 24 algebra (1 + t)3 = 1 + 3t + 3t 2 + t 3 (sec2 x 1) sec3 x(sec x tan x) dx (v 2 1)v 3 dv = (v 5 v 3 ) dv tan3 x(1 + tan2 x)(sec2 x) dx u3 (1 + u2 ) du = (u3 + u5 ) du
u6 tan4 x tan6 x u4 + +C = + +C 4 6 4 6
v4 sec6 x sec4 x v6 +C = +C 6 4 6 4
TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS
[CHAP. 39
= = = and so the two expressions for
4(1 + 3u2 + 3u4 + u6 ) 6(1 + 2u2 + u4 ) 24 4 + 12u2 + 12u4 + 4u6 6 12u2 6u4 24 6u4 + 4u6 2 u4 u6 1 = + 24 4 6 12
1 tan3 x sec4 x dx are equivalent. (The 12 is soaked up by the arbitrary constant C.)
39.6 Find
tan2 x sec x dx.
Problem 39.5 is of no help here. tan2 x sec x dx = = (sec2 x 1) sec x dx = (sec3 x sec x) dx [by Problem 38.13(b)]
1 (sec x tan x + ln |sec x + tan x|) ln |sec x + tan x| + C 2 1 1 = sec x tan x ln |sec x + tan x| + C 2 2
39.7 Prove the trigonometric identity sin Ax cos Bx = 1 (sin(A + B)x + sin(A B)x). 2
The sum and difference formulas of Theorem 26.6 give sin (A + B)x = sin (Ax + Bx) = sin Ax cos Bx + cos Ax sin Bx sin (A B)x = sin (Ax Bx) = sin Ax cos Bx cos Ax sin Bx and so, by addition, sin (A + B)x + sin (A B)x = 2 sin Ax cos Bx.
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