What is the integral of xsin2xdx?

Answer 1

∫ xsin(2x) dx = (-1/2)xcos2x + (1/4)sin2x

You get this by using Integration by Parts.

An integral in the form ∫udv can be written as uv-∫vdu

In the case of your problem u=x, du=1, dv=sin2x, v=(-1/2)cos2x <--You get v by integrating dv

Using the formula ∫udv = uv- ∫vdu and by plugging in what has been defined above you get ∫xsin(2x)dx = (-1/2)xcos2x - ∫(-1/2)cos2x(1)

By integrating ∫(-1/2)cos2x, you get (-1/4)sin2x.

When you plug that back in, you get ∫xsin2xdx=(-1/2)xcos2x-(-1/4)sin2x or just simply

∫xsin(2x)dx = (-1/2)xcos(2x) + (1/4)sin(2x)