TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS
239
29.9
29.10
29.11
29.12
29.13
29.14
29.15
29.16
29.17
Use Problem 28.39:
Since the exponent of tan x is odd, f tan
3 x sec
3 dx = J (sec
2 x - 1) sec
2 x sec x tan ;e dx = f (sec
4 * sec x
By Problems 28.39 and 28.40,
and
Then
Use the formula
So
So
Recall
cos >l;e cos Bx = |[cos (.4 - B)JC + cos (A + B)x]
f tan
2 x sec
4 * tie.
f tan
3 x sec
3 x dx.
Since the exponent of sec x is even, / tan
2 x sec
4 x dx = J tan
2 je (1 + tan
2 A:) sec
2 x dx = J (tan
2 x sec
2 * +
tan
4 x sec
2 x) tie = 5 tan
3 x + I tan
5 x + C.
/ sec
5 je tie.
tan x - sec2 AC sec x tan *) tie = \ sec5 * - 5 sec3 x + C.
/ tan
4 * sec x dx.
/ tan
4 x sec x tie = /(sec
2 x- I)
2 sec * dx = f (sec
4 * - 2 sec
2 * + 1) sec x dx = f (sec
5 x - 2 sec
3 x +
sec *) dx.
/ sec
5 xdx =
sec
3 * tie,
/ sec
3 A: dx =
Thus, we get
sec
3 x dx — 2 J sec
3 A: tie + In jsec * +
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
= sin
2 AC cos
2 x + C.
J sin irx cos 3 me tie.
sin Ac cos fi* = i[sin (A + B)x + sin (,4 - B)x].
J sin TTX cos3irx dx =
| J [sin 4-rrx + sin (—2trx)] dx = \ J (sin 47r;e — sin 2
cos 47rje) + C.
J sin 5x sin Ix dx.
Recall
sin Ax sin fte = 5[cos (A - B)x - cos (^4 + B)x].
J sin 5x sin ?A; tie = \ / [cos (-2*) -
cos 12jc] dx=\l (cos 2x - cos \2x) dx =
$ cos 4x cos 9 xdx.
J cos 4x cos 9* tie = \ J [cos (-5x) +
cos 13*] tie = 5 J (cos 5x + cos 13x) tie =
Calculate J J sin nx sin Aa: tie when n and A: are distinct positive integers.
So
sin nx sin kx = \ [cos (n - k)x - cos (n + k)x]. Jp" sin nx sin fce tie = | J0" [cos (n - k)x - cos (« + k)x] dx
(6 sin 2x - sin 12x) + C.