integrate sin^(3)2xcos^(2)2xdx?

Answer 1

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

First, break off a cos(2x).

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

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

Integral ( sin^3(2x) (1 - sin^2(2x)) cos(2x) dx)

Now, use substitution.

Let u = sin(2x). Then
du = 2cos(2x) dx, so
(1/2) du = cos(2x) dx.

Note that the tail end of our integral is cos(2x) dx, so the tail end after the substitution will be (1/2) du.

Integral ( u^3 (1 - u^2) (1/2) du )

Pull the constant (1/2) out,

(1/2) Integral ( u^3 (1 - u^2) du )

Expand.

(1/2) Integral ( (u^3 - u^5) du )

Now, integrate using the reverse power rule.

(1/2) [ (1/4)u^4 - (1/6)u^6 ] + C

Distribute the (1/2),

(1/8)u^4 - (1/12)u^6 + C

Substitute back u = sin(2x),

(1/8)sin^4(2x) - (1/12)sin^6(2x) + C