There's a few ways to tackle this problem, but personally I would expand the integrand by using various trigonometric identities, including the product-to-sum identities and double-angle formulas.
Here are the ones I used specifically for this problem:
sinalphacosbeta=1/2[sin(alpha-beta)+sin(alpha+beta)]sinαcosβ=12[sin(α−β)+sin(α+β)]
sinalphasinbeta=1/2[cos(alpha-beta)-cos(alpha+beta)]sinαsinβ=12[cos(α−β)−cos(α+β)]
sin2alpha=2sinalphacosalphasin2α=2sinαcosα
So, let's use them. You can pick any pair. I picked sinthetasin2thetasinθsin2θ first:
int(sinthetasin2theta)sin3thetad theta∫(sinθsin2θ)sin3θdθ
=int(1/2)(costheta+cos3theta)(sin3theta)d theta=∫(12)(cosθ+cos3θ)(sin3θ)dθ
=(1/2)intsin3thetacostheta-sin3thetacos3thetad theta=(12)∫sin3θcosθ−sin3θcos3θdθ
=(1/2)int(1/2)(sin4theta+sin2theta)-(1/2)sin6thetad theta=(12)∫(12)(sin4θ+sin2θ)−(12)sin6θdθ
=(1/4)int(sin4theta+sin2theta-sin6theta)d theta=(14)∫(sin4θ+sin2θ−sin6θ)dθ
At this point, it's a simple indefinite integral:
(1/4)[-cos(4theta)/4-cos(2theta)/2+cos(6theta)/6]+C(14)[−cos(4θ)4−cos(2θ)2+cos(6θ)6]+C
=cos(6x)/24-cos(4x)/16-cos(2x)/8+C=cos(6x)24−cos(4x)16−cos(2x)8+C