I'm trying to solve the integral ∫0π23+sin⁡(2θ)dθI know the initial trigonometric substitution changes the integral to ∫0π2z2+6iz−1dzand we use the substitutions dθ=dziz and sin⁡(2θ)=z2−z−22i to get there.My best attempt is reworking the integral denominator to 6i+z2−1z2(iz)

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I&#039;m trying to solve the integral ∫0π23+sin⁡(2θ)dθI know the initial trigonometric substitution changes the integral to ∫0π2z2+6iz−1dzand we use the substitutions dθ=dziz  and  sin⁡(2θ)=z2−z−22i to get there.My best attempt is reworking the integral denominator to 6i+z2−1z2(iz)

Ethan Sanders · Accepted Answer

It looks like you&#039;ve actually substituted z=e2iθ so that 2isin⁡θ=z−z−1, giving the contour integral ∮2dz/(2iz)3+(z−z−1)2i=∮2dzz2+6iz−1 The denominator is (z+3i)2+8, so the poles are (−3±22)i, both first-order. However, the only one that contributes by the residue theorem is the one within the |z|=1 contour, (−3+22)i. This has residue 122i, so the integral is 2πi22i=π2

I'm trying to solve the integral \int_0^{\pi} \frac{2}{3+\sin(2 \theta)}d \theta I

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