Prove that $\int_{-\pi}^{\pi}$ $\frac{d\theta}{1+\sin^2\theta}$ = $\pi\sqrt{2}$ using the method of Residues Prove that $$\int_{-\pi}^{\pi}\frac{d\theta}{1+\sin^2\theta} = \pi\sqrt{2}$$ using the method of Residues
How do I do this? I know I need it from $0$ to $2\pi$ but I don't know how to modify it!!
Help me!
 A: Note: there was a computation error; this is now fixed. I'm surprised no one noticed this.
Why don't you start by writing $z = e^{i\theta}$ and use the fact that $\sin(\theta) = 1/
(2i)\cdot(e^{i\theta} - e^{-i\theta}) = \frac{z-1/z}{2i}$ and then you are integrating $\int_{|z|=1} \frac{-4 dz}{(iz)\cdot(-4+z^2-2+1/z^2)} = \int_{|z|=1} \frac{4iz dz}{z^4-6z^2+1}$. Now, solve using the Residue Theorem.
Additional part of question:
What are the roots of $z^4-6z^2 + 1$? Note that $z^2 = 3 +- 2\sqrt{2}$. Now, $3 + 2\sqrt{2} > 1$ and so we can ignore this, since $\sqrt{1} = 1$. I leave it to you to figure out why. So we consider the roots when $z^2 = 3 - 2\sqrt{2}$. Then, we can factor the denominator as $(z^2 - 3 - 2\sqrt{2})(z^2 - 3 + 2\sqrt{2})$. Now, we add the two residues which gives $\frac{\sqrt{3-2\sqrt{2}}}{2\sqrt{3-2\sqrt{2}}\cdot (-4\sqrt{2})} + \frac{-\sqrt{3-2\sqrt{2}}}{-2\sqrt{3-2\sqrt{2}}\cdot(-4\sqrt{2})} = -1/(4\sqrt{2})$. Now, we multiply by the $4i$ in the numerator to give $-i/\sqrt{2}$. Then, multiply by $2\pi i$ to give an answer of $\pi\sqrt{2}$. 
