Really weird/hard integration calculus problem?

Question

Integrate the following indefinite integral:

(sin7x)^12 * (cos7x)^3

Hint: sin^2 + cos^2 = 1

Answer 1

Not very hard once you know the trick!
∫sin^12 (7x) cos^3 (7x) dx
= ∫sin^12 (7x) (1 - sin^2 (7x)) cos (7x) dx
Substitute u = sin 7x, du = 7 cos 7x dx.
= ∫u^12 (1 - u^2) (1/7) du
= (1/7) ∫(u^12 - u^14) du
= (1/7) (u^13 / 13 - u^15 / 15) + c
= (1/7) (sin^13 (7x) / 13 - sin^15 (7x) / 15) + c.

In general, to integrate sin^n x cos^m x, if n or m or both are odd, pick the lowest odd one; the derivative should be part of this term, so pick the OTHER function to substitute.
e.g. sin^3 x cos^5 x: lowest odd power is 3, so substitute u = cos x.
sin^2 x cos^7 x: lowest odd power is 7, so substitute u = sin x.

If n and m are both even, use the double-angle formulae to rewrite it in terms of sin 2x and cos 2x: e.g.
sin^4 x cos^6 x = (sin x cos x)^4 cos^2 x
= ((1/2) sin 2x)^4 ((1/2) (cos 2x + 1))
= 1/32 (sin^4 (2x) cos (2x)) + 1/32 sin^4 2x
The first part of this can be dealt with by using u = sin (2x) as before; for the second part we use the double angle formulae again to rewrite
sin^4 2x = (sin^2 2x)^2 = ((1/2) (1 - cos 4x))^2
= 1/4 (1 - 2 cos 4x + cos^2 4x)
of which the first two terms can be integrated directly and we do the same thing yet again on the final term to get an integral involving cos 8x.

Answer 2

Try this:

1) let a = 7x, and rewite as

(1/7) int(sin^12 a cos^3 a) da

2) rewrite as

(1/7) int(cos^2 a sin^12 a cos a) da, int by parts

let u = cos^2 a
du = 2 cos a (-sin a) da
= -2 cos a sin a da

dv = sin^12 a cos a da, and let

w = sin a. Note then that

dv = w^12 dw or

v = w^13/13 or

v = sin^13 a /13

So the integral we want is now

(1/7) [ uv - int(v du)] or

(1/7) [ cos^2 a * sin^13 a/13 - int( sin^13/13 * (-2 cos a sin a))]

This last integral can be simplified to

+ (2/13) int(sin^14 a cos a da) or

(2/13) sin^15 a/15, so final answer looks like

(1/7)[cos^2 a * sin^13 a /13 + (2/(13*15)) sin^15 a ]

Clean this up.

Answer 3

∫(sin7x)^12 * (cos7x)^3 dx
= ∫(sin7x)^12 * (cos7x)^2 (cos7x) dx
= ∫(sin7x)^12 *(1- (sin7x)^2) (cos7x) dx
Let u = sin(7x)
du = 7cos(7x)dx
∫(sin7x)^12 *(1- (sin7x)^2) (cos7x) dx
= (1/7)∫u^12 (1-u^2) du
= (1/7)∫u^12 - u^14 du
= (1/7)((u^13)/13 - (u^15)/15)) + c

= (1/91)(sin7x)^13 - (1/105)(sin7x)^15 + c

Answer 4

I think it's this: Integral (sin7x)^12 * (cos7x)^3 cos^2 = 1 - sin^2 So subbing that identity gives you: Integral (sin7x)^12 * (cos7x)*(1 - sin^2) Split it up to: Integral (sin7x)^12 * (cos7x) + Integral - (sin7x)^14 * (cos7x) and then take u-subs?

Answer 5

This problem has too many steps to show out. The answer is:
(99sin7x)/114688
-(187sin21x)/344064
+(99sin35x)/573440
+(3sin49x)/114688
-(19sin63x)/344064
+(3sin77x)/114688
-(9sin91x)/1490944
+sin(105x)/1720320
This was checked by http://integrals.wolfram.com/index.jsp