Question

integral of sin^2(x)cos(3x)dx

Answer 1

Find cos(3x) in terms of cos(x).
cos(3x) = cos(2x + x) = ?

Answer 2

Thanks, its cos(2x)cos(x)-sin(2x)sin(x), right?
If so I think i got it from here

Answer 3

Correct. Go further and expand cos(2x) and sin(2x).

Answer 4

I got that to become 4cos^3(x)-3cos(x) which then results in int sin^2(x)4cos^3(x) -int sin^2(x)cos(x)

Answer 5

sin^2(x) * { 4cos^3(x) - cos(x) } =
sin^2(x) * { 4cos^2(x) - 1 } * cos(x) =
sin^2(x) * { 4(1-sin^2x) - 1 } * cos(x) =
sin^2(x) * { 3 - 4sin^2x } * cos(x) =
Let u = sin(x)
u^2 * (3 - 4u^2) * du

Answer 6

I have not checked if cos(3x) = 4cos^3(x) - cos(x) that you got is correct.

Answer 7

I believe it is, i just finished the problem and everything worked out so hopefully that was correct
Thanks a lot

Answer 8

You are welcome.