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Trigonometric Integrands and Substitutions solved questions
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CHAPTER 29Trigonometric Integrands
and Substitutions
29.1 Find J cos2 ax dx.
29.2 Find / sin2 ax dx.
Using Problem 29.1, J sin2 ax dx =
In Problems 29.3-29.16, find the indicated antiderivative.
29.3 J sin x cos2 x dx.
Let M = cos x, du = —sin x dx.Then J sin x cos2 dx = - J u2 du = - i«3 + C = - \ cos3 x + C.
29.4 J sin4 x cos5 x dx.
Since the power of cos* is odd, let « = sin;e, du = cos x dx. Then Jsin 4 A: cos5 x dx = J sin4 x(l —
sm2x)2cosxdx = $u\l-u2)2 du = J «4(1 - 2«2 + u4) du = J (u4 - 2u6 + «8) du = ^w5 - §w7 + \u" + C =
w5(i - |«2 + |a4) + C = sin5 x($ - f sin2 * + | sin4 x) + C.
29.5 J cos6 x dx.
$cos6xdx = J (cos2 x)3 <& = (1 + 3 cos 2x + 3 cos2 2x + cos3 2*) dx =
Now,
Also, in J(l-sin 2 2jt)cos2;edx, let u = sin 2x, du=2 cos 2* djt. So we get
Hence, the entire answer is
[Can
you show that this answer agrees with Problem 28.36?
29.6 J cos 4 x sin2 x dx.
J cos4 x sin2 x dx =
cos 2x - cos 2x - cos 2*) dx = I (x + | sin 2x - J cos 2x dx - J cos 2x dx). Now, J cos 2x dx = \ (x +
\ sin 2x cos 2x) by Problem 29.1. Also, J cos3 2x dx = / (1 - sin2 2x) cos 2x dx = / cos 2x dx -
J sin2 2* cos 2x dx = | sin 2x — \ sin3 IK. Hence, we get \ [x + \ sin 2x — \ (x + | sin 2x cos 2x) + \ sin 2x -
g sin3 2x] + C = s [(x/2) + sin 2x - \ sin 2x cos 2x - g sin3 2*] + C.
29.7
Let * = 2«, <& = 2 d«. Then
29.8 J tan4 x dx.
J tan4 x dx = / tan 2 * (sec2 * - 1) dx = J tan 2 x sec2 x <& - J tan2 x dx = j tan3 x - J (sec2 x - 1) dx
= 3 tan3 x - tan x + x + C
238
, TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 239
29.9 / sec5 je tie.
Use Problem 28.39:
29.10 f tan2 x sec4 * tie.
Since the exponent of sec x is even, / tan2 x sec4 x dx = J tan2 je (1 + tan2 A:) sec2 x dx = J (tan2 x sec2 * +
tan x sec2 x) tie = 5 tan3 x + I tan5 x + C.
4
29.11 f tan3 x sec3 x dx.
Since the exponent of tan x is odd, f tan3 x sec3 dx = J (sec2 x - 1) sec2 x sec x tan ;e dx = f (sec4 * sec x
tan x - sec2 AC sec x tan *) tie = \ sec5 * - 5 sec3 x + C.
29.12 / tan4 * sec x dx.
/ tan4 x sec x tie = /(sec2 x - I)2 sec * dx = f (sec4 * - 2 sec2 * + 1) sec x dx = f (sec5 x - 2 sec3 x +
sec *) dx. By Problems 28.39 and 28.40, / sec5 xdx = sec3 * tie, and / sec3 A: dx =
Thus, we get sec3 x dx — 2 J sec3 A: tie + In jsec * +
29.13 J sin 2x cos 2x dx.
/ sin 2* cos 2x dx = | J sin 2x • 2 cos 2x dx = \ • | sin2 2x+C=\ sin2 2x + C = \(2 sin x cos x)2 + C
= sin2 AC cos2 x + C.
29.14 J sin irx cos 3 me tie.
Use the formula sin Ac cos fi* = i[sin (A + B)x + sin (,4 - B)x]. Then J sin TTX cos3irx dx =
| J [sin 4-rrx + sin (—2trx)] dx = \ J (sin 47r;e — sin 2
cos 47rje) + C.
29.15 J sin 5x sin Ix dx.
Recall sin Ax sin fte = 5[cos (A - B)x - cos (^4 + B)x]. So J sin 5x sin ?A; tie = \ / [cos (-2*) -
cos 12jc] dx=\l (cos 2x - cos \2x) dx = (6 sin 2x - sin 12x) + C.
29.16 $ cos 4x cos 9 xdx.
Recall cos >l;e cos Bx = |[cos (.4 - B)JC + cos (A + B)x] So J cos 4x cos 9* tie = \ J [cos (-5x) +
cos 13*] tie = 5 J (cos 5x + cos 13x) tie =
29.17 Calculate J J sin nx sin Aa: tie when n and A: are distinct positive integers.
So sin nx sin fce tie = | J0" [cos (n - k)x - cos (« + k)x] dx
sin nx sin kx = \ [cos (n - k)x - cos (n + k)x]. Jp"
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