Question

How do you find the integral of `int [sin(3x)]^4 dx`?

Answer 1

Use power reduction formulas twice.

Explanation:

`cos(2x) = cos^2x-sin^2x`
`cos(2x) = 1-2sin^2x`
`cos(2x) = 2cos^2x-1`

So,
`sin^2x = 1/2(1-cos(2x))`
and,
`cos^2x = 1/2(1+cos(2x))`

`[sin(3x)]^4 = [color(blue)(sin^2 3x)]^2`

`= [color(blue)(1/2(1-cos6x))]^2`

`= 1/4[1-2cos6x+color(red)(cos^2 6x)]`

`= 1/4[1-2cos6x+(color(red)(1/2(1+cos12x)))]`

`= 1/8[2(1-2cos6x)+(1+cos12x))]`

`= 1/8[3-4cos6x+cos12x]`

The last expression can be integrated term by term using substitution for the two terms involving cosine.