please help evaluate the integral

Question

Help me evaluate this integral. i don't know where to start and have been stressed over it

Answer 1

Integrate by parts. Let u=θ² and dv=sin(2θ)dθ, so:

du=2θdθ, v=-cos(2θ)/2.

∫udv=uv-∫vdu=-θ²cos(2θ)/2+∫θcos(2θ)dθ.

∫θcos(2θ)dθ by parts: let u=θ, du=dθ; dv=cos(2θ)dθ, v=sin(2θ)/2.

∫udv=θsin(2θ)/2-½∫sin(2θ)dθ=θsin(2θ)/2+¼cos(2θ).

Complete integral is:

-θ²cos(2θ)/2+θsin(2θ)/2+¼cos(2θ)+C where C is integration constant.

CHECK

Differentiate -θ²cos(2θ)/2+θsin(2θ)/2+¼cos(2θ)+C:

-θcos(2θ)+θ²sin(2θ)+½sin(2θ)+θcos(2θ)-½sin(2θ)=θ²sin(2θ). Checks out OK.