I'm trying to solve the integral $\int_0^\pi \frac{2}{3+sin(2\theta)}d\theta$.
I know the initial trigonometric substitution changes the integral to $\int_0^\pi \frac{2}{z^2+6iz-1}dz$
and we use the substitutions $d\theta=\frac{dz}{iz}$ and $sin(2\theta)=\frac{z^2 - z^{-2}}{2i}$ to get there.
My best attempt is reworking the integral denominator to $6i+z^2-\frac{1}{z^2}(iz)$. May I please have some assistance?
On
Why not to substitute $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$ to get to answer $$\int_0^{\pi} \frac{2d\theta}{3+\sin2\theta}$$ $$=2\int_0^{\pi/2} \frac{2d\theta}{3+\sin2\theta}$$ $$=4\int_0^{\pi/2} \frac{d\theta}{3+\frac{2\tan\theta}{1+\tan^2\theta}}$$ $$=4\int_0^{\pi/2} \frac{\sec^2\theta d\theta}{3\tan^2\theta+2\tan\theta+3}$$ $$=\frac43\int_0^{\pi/2} \frac{\sec^2\theta d\theta}{\left(\tan\theta+\frac13\right)^2+\frac89}$$ $$=\frac43\int_0^{\pi/2} \frac{d\left(\tan\theta+\frac13\right)}{\left(\tan\theta+\frac13\right)^2+\left(\frac{2\sqrt2}{3}\right)^2}$$ $$=\sqrt2\left(\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2\sqrt2}\right)\right)$$
It looks like you've actually substituted $z=e^{2i\theta}$ so that $2i\sin\theta=z-z^{-1}$, giving the contour integral$$\oint\frac{2dz/(2iz)}{3+(z-z^{-1})/2i}=\oint\frac{2dz}{z^2+6iz-1}.$$ The denominator is $(z+3i)^2+8$, so the poles are $(-3\pm2\sqrt{2})i$, both first-order. However, the only one that contributes by the residue theorem is the one within the $|z|=1$ contour, $(-3+2\sqrt{2})i$. This has residue $\frac{1}{2\sqrt{2}i}$, so the integral is $\frac{2\pi i}{2\sqrt{2}i}=\frac{\pi}{\sqrt{2}}$.