Typically, if you have an odd power of sine or cosine you can factor one out, to use with the "dx", then use sin2(x)+cos2(x)=1, so that either sin2(x)=1−cos2(x) or cos2(x)=1−sin2(x) to reduce the remaining even power. Here, you have ∫0π/2sin3(θ)cos2(θ)dθ which we can write as ∫0π/2sin2(θ)cos2(θ)(sin(θ)dθ)=∫0π/2(1−cos2(θ))cos2(θ)(sin(θ)dθ)=∫0π/2(cos2(θ)−cos4(θ))(sin(θ)dθ).
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