# &int; xSin(x)Sin(2x) dx

A few weeks ago, I saw a post regarding the area under the function xsinxsin2x dx, (x = pi*x/a). After cogitations, I have found the answer:

&int;x*Sin(x)*Sin(2x) dx = - xSin(2x)Cos(x) + &int;Cos(x) dx
= Sin(x) - xSin(2x)Cos(x) + C

By assuming u = xsin2x, du = 2xcos2x + sin2x, dv = sinx, v = -cosx. And using the Parts formula.

## Answers and Replies

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
The IBP formula is

&int; u dv = uv - &int; v du

not &int; v dx

BTW, the key to this is the trig identity
2 sin x cos x = sin 2x