What is the solution to the integral of xSin(x)Sin(2x) dx?

[PrudensOptimus]

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:

∫x*Sin(x)*Sin(2x) dx = - xSin(2x)Cos(x) + ∫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.

 

[Hurkyl]

The IBP formula is

∫ u dv = uv - ∫ v du

not ∫ v dx


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

 

[M98Ranger]

The solution to the integral of xSin(x)Sin(2x) dx is Sin(x) - xSin(2x)Cos(x) + C. This was found by using the integration by parts formula and assuming u = xsin2x and dv = sinx. After solving for du and v, the formula was used to find the solution. It is important to note that the constant C was added to account for the indefinite integral. This solution can be verified by taking the derivative, which will result in the original function.