I would like to solve the following integral using Cauchy Integral Theorem. I tried, but I got a different solution to Wolfram Alpha (Wolframs solution is pi/60). Help please!
$$\int_{0}^{2 \pi} {{\sin \theta} \over {34 - 16 \sin \theta}} d\theta$$
This is how I attempted the Cauchy Integral Theorem:
let $$ z =e^{i\theta} $$ $$\sin \theta = {1 \over 2i}(z - {1 \over z})$$ $$d\theta = {1 \over iz} dz$$ I proceed to substitute sin theta with the function of z and get the following expression, skipping ahead to get the following $${i \over 2} \oint_{c}{{z - {1 \over z}} \over(8z^2 - 34iz-8)} dz$$ $${i \over 2} \oint_{c}{{z - {1 \over z}} \over(z -4i)(z -0.25i)} dz$$ I then applied the theorem to the only pole in the unit circle, Z - 0.25 i and got the answer
$$17 \pi \over 15$$ Where did I do wrong? I've ran through the workings countless times but I cant seem to get the answer.
Ah! This is great stuff! Gotta love Cauchy.
I agree with all of your work except your statement that the only pole is at $z=0.25i$. If you factor the top, you get an expression like $$z-\frac{1}{z}=\frac{z^2-1}{z}$$ which gives you a pole also at $z=0$.
$$f(z)=\frac{i}{2}\frac{(z+1)(z-1)}{(z)(8z^2-34iz-8)}$$
Adding up the residues using Cauchy's Residue Theorem, \begin{align*} \int_0^{2\pi}\frac{\sin{\theta}}{34-16\sin{\theta}}d\theta&=2\pi i\left(\text{Res}(f,z=0)+\text{Res}(f,z=0.25i)\right)\\ &=2\pi i\left(\frac{i}{16}-\frac{17i}{240}\right)\\ &=\frac{\pi}{60} \end{align*}