i've gotten stuck on a certain integral and i could appreciate the help.
I want to integrate $$\int_{0}^{\pi}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta},~\alpha > 0$$
now i believe the answer is $$\frac{\pi}{\sqrt{a^2 + 1}}$$ (via, Wolfram) but as it currently stands i have no idea how to get there, The only stipulation is that i am required to use the residue theorem. so...heres what i've tried
Attempt one: Substitute $\sin \theta = \frac{1}{2i}\left(z+\frac{1}{z}\right)$ \begin{align*} \int_{0}^{\pi}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} &= \\ \alpha\int_{\gamma}\frac{iz~dz}{\alpha^2+\frac{1}{4}\left(z+\frac{1}{z}\right)^{2}} &= \\ \frac{\alpha}{4}\int_{\gamma}\frac{iz~dz}{4\alpha^2+\left(\frac{z^4+2z^2+1}{z^2}\right)} &= \\ \frac{\alpha}{4}\int_{\gamma}\frac{z^2\cdot iz~dz}{z^4+z^2(4\alpha^2+2)+1} &= \\ \end{align*}
but at this point i admit i gave up given that i ended up with something kinda ugly, so i believe theres a smarter way of proceeding.
Attempt Two: Substitute $\theta = \tan^{-1}(t)$ \begin{align*} \int_{0}^{\pi}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} &= \\ \alpha~\int_{0}^{\pi}\frac{d\theta}{\alpha^2+\sin^{2}(\tan^{-1}(t))} &= \\ \alpha~\int_{}^{}\frac{t^2+1}{\alpha^2(t^2+1)-t^2} \frac{dt}{t^2+1} &= \\ \alpha~\int_{}^{}\frac{dt}{\alpha^2(t^2+1)-t^2} &= \\ \end{align*} the issue i'm having here is the bounds of integration. since we're originally integrating between 0 and pi, i would throw that into $\tan(\theta)=t$ but $\tan{0}=0 = \tan{\pi}$ so i'm stuck here too. and whats worse is the denominator doesnt really look like its going to factor into anything especially useful.
as it currently stands im just uncertain how to integrate this type of function, ie should i treat it as a trig function $Q(cos\theta,sin\theta)$, introducing substitutions for cos and sin then integrating along the contour of $|z|=1$? or should i be trying to get it into some form where the bounds of integration is the real axis and then using the upper half of the plane?
So...any suggestions would be greatly appreciated.
The complex analysis approach can be simplified substantially if power reduction formula is applied.
In particular, $$\sin^2\theta=\frac{1-\cos 2\theta}{2}$$
Hence, $$\begin{align} \int_{0}^{\pi}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} &=\int_{0}^{\pi}\frac{2\alpha~d\theta}{2\alpha^2+1-\cos 2\theta} \\ &\stackrel{u=2\theta}{=}\int_{0}^{2\pi}\frac{\alpha~du}{2\alpha^2+1-\cos u} \\ &\stackrel{z=e^{iu}}{=}\oint_{|z|=1}\frac{2\alpha}{4\alpha^2+2-z-z^{-1}}\frac{dz}{iz} \\ &=\oint_{|z|=1}\frac{2i\alpha}{z^2-(4\alpha^2+2)z+1}dz \\ &=2i\alpha\oint_{|z|=1}\frac{1}{(z-r_+)(z-r_-)}dz \end{align} $$
Here comes the most intensive part of the whole calculation: finding the roots. After a little algebra, we get $$r_{\pm}=2\alpha^2+1\pm 2\alpha\sqrt{\alpha^2+1}$$
Clearly, $|r_+|>1, |r_-|<1$. Therefore, by residue theorem,
$$\begin{align} \int_{0}^{\pi}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} &=2i\alpha\oint_{|z|=1}\frac{1}{(z-r_+)(z-r_-)}dz \\ &=2i\alpha\cdot2\pi i\operatorname*{Res}_{z=r_-}\frac{1}{(z-r_+)(z-r_-)} \\ &=-4\pi\alpha\cdot\frac1{r_--r_+} \\ &=\frac{-4\pi\alpha}{-2\cdot2\alpha\sqrt{\alpha^2+1}} \\ &=\color{red}{\frac{\pi}{\sqrt{\alpha^2+1}}} \end{align} $$
Regarding attempt 2, whenever you are doing u-substitution on $\displaystyle{\int^b_a(\cdot)dx}$, always make sure that the substitution is of the form $$u=\text{continuous injective function}(x)$$ (continuous and injective on $(a,b)$).
Clearly $u=\tan x$ does not satisfy this condition.
However, this does not bar $u=\tan x$ from solving the integral. Simply break the original into two part before doing the substitution: $$\begin{align} \int_{0}^{\pi}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} &=\int^{\pi/2^{-}}_0 \frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} +\int^{\pi}_{\pi/2^{+}}\frac{\alpha~d\theta}{\alpha^2+\sin^{2}\theta} \\ &=\alpha~\int^{\infty}_{0}\frac{dt}{\alpha^2(t^2+1)-t^2}+\alpha~\int_{-\infty}^{0}\frac{dt}{\alpha^2(t^2+1)-t^2} \\ &=\alpha~\int^{\infty}_{-\infty}\frac{dt}{(\alpha^2-1)t^2+\alpha^2} \\ &\stackrel{v=t\sqrt{\alpha^2+1}}{=}\frac{\alpha}{\sqrt{\alpha^2+1}}\cdot\int^{\infty}_{-\infty}\frac{dv}{v^2+\alpha^2} \\ &\stackrel{w=v/\alpha}{=}\frac{1}{\sqrt{\alpha^2+1}}\cdot\int^{\infty}_{-\infty}\frac{dw}{w^2+1} \end{align} $$ and the rest should be straight-forward.