Indefinite Integral?

Question

sin^3(12x)cos^2(12x)dx

Answer 1

∫sin³ (12x) cos² (12x) dx

First, extract a factor of sin² (12x):

∫sin (12x) sin² (12x) cos² (12x) dx

Write sin² (12x) as 1-cos² (12x):

∫sin (12x) (1-cos² (12x)) cos² (12x) dx

Expand:

∫sin (12x) cos² (12x) - sin (12x) cos⁴ (12x) dx

Let u=cos (12x), du=-12 sin (12x) dx, so that:

1/12 ∫u⁴ - u² du

Integrate:

u⁵/60 - u³/36 + C

Resubstitute:

cos⁵ (12x)/60 - cos³ (12x)/36 + C

And we are done.

Edit: corrected sign error.

Answer 2

Integral ( sin^3(12x) cos^2(12x) dx )

First, break off a sin(12x).

Integral ( sin^2(12x) cos^2(12x) sin(12x) dx )

Use the identity sin^2(z) = 1 - cos^2(z).

Integral ( [1 - cos^2(12x)] cos^2(12x) sin(12x) dx )

Use substitution.
Let u = cos(12x). Then
du = -12sin(12x) dx, which means
(-1/12) du = sin(12x) dx

Integral ( [1 - u^2] u^2 (-1/12) du )

Pull the constant out of the integral.

(-1/12) Integral ( [1 - u^2] u^2 du )

Expand by distributing u^2.

(-1/12) Integral ( (u^2 - u^4) du )

And now use the reverse power rule.

(-1/12) [ (1/3)u^3 - (1/5)u^5 ] + C

Distribute the -1/12,

(-1/36)u^3 + (1/60)u^5 + C

But u = cos(12x), so our final answer is

(-1/36)cos^3(12x) + (1/60)cos^5(12x) + C