Evaluate $\int \sqrt{ \frac {\sin(x-\alpha)} {\sin(x+\alpha)} }\,\operatorname d\!x$? How to go about evaluating the following integral?
$$ I = \int \sqrt{ \dfrac {\sin(x-\alpha)} {\sin(x+\alpha)} }\,\operatorname d\!x$$
What I have done so far:
$$ I = \int \sqrt{ 1-\tan\alpha\cdot\cot x }\,\operatorname d\!x$$
Let $ t^2 = 1-\tan\alpha\cdot\cot x $
$$ \begin{align} 
 2t\,\operatorname d\!t &= \tan\alpha \cdot \csc^2x\,\operatorname d\!x \\
 & = \tan\alpha \cdot \Bigg(1 + \Big(\dfrac{1-t^2}{\tan\alpha}\Big)^2\Bigg)\,\operatorname d\!x \\
 & = \dfrac{\Big(\tan^2 \alpha + (1-t^2)^2\Big)}{\tan \alpha}dx \end{align}$$
So, from that:
$$\begin{align}
 I &= \int \sqrt{ 1-\tan\alpha\cdot\cot x }\,dx \\
 & = \int \dfrac{2t^2\tan\alpha}{\Big(\tan^2 \alpha + (1-t^2)^2\Big)}\, \operatorname d\!t \\
\end{align}$$
What to do next?
Edit:
I had thought of doing a substitution: $u = 1-t^2$ but that doesn't work as you need one more $t$ term in the numerator.
 A: Substitute
$$u = \frac{\sin{(x-\alpha)}}{\sin{(x+\alpha)}}$$
Then, with some algebraic manipulation, we find that
$$dx = \frac{2 du}{\sec^2{\alpha} u^2 + 2 (\tan^2{\alpha}-1) u + \sec^2{\alpha}}$$
so that the integral becomes
$$2 \int du \frac{\sqrt{u}}{\sec^2{\alpha} \, u^2 + 2 (\tan^2{\alpha}-1) u + \sec^2{\alpha}}$$
As for the latter integral, break up into its factors $u-u_{\pm}$, where
$$u_{\pm} = \cos{2 \alpha} \pm i \cos{\alpha}$$
and do a partial fractions decomposition, so the integral becomes
$$\frac{1}{i 2 \cos{\alpha}} \left [ \int du \frac{\sqrt{u}}{u-u_+} - \int du \frac{\sqrt{u}}{u-u_-}\right ]$$
To evaluate each of these integrals, let $u=v^2$ so that
$$\int du \frac{\sqrt{u}}{u-u_+} = 2 \int dv \frac{v^2}{v^2-u_+} = 2 v + 2 u_+ \int \frac{dv}{v^2-u_+}$$
the latter integral taking the form of an inverse hyperbolic tangent.  The result I get is
$$\int dx \sqrt{\frac{\sin{(x-\alpha)}}{\sin{(x-\alpha)}}} = \frac{1}{\cos{\alpha}} \Im{\left [\sqrt{u_+} \log{\left ( \frac{\sqrt{u} - \sqrt{u_+}}{\sqrt{u} + \sqrt{u_+}}\right)}\right]} + C$$
where, again
$$u = \frac{\sin{(x-\alpha)}}{\sin{(x-\alpha)}} $$
$$u_{+} = \cos{2 \alpha} + i \cos{\alpha}$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
I&\equiv\int\root{\sin\pars{x - \alpha} \over \sin\pars{x + \alpha}}\,\dd x
=\int\root{\sin\pars{x}\cos\alpha - \cos\pars{x}\sin\pars{\alpha}\over
\sin\pars{x}\cos\alpha + \cos\pars{x}\sin\pars{\alpha}}\,\dd x
\\[3mm]&=\int\root{\tan\pars{x} - \beta\over
\tan\pars{x} + \beta}\,\dd x\quad\mbox{where}\quad\beta \equiv\tan\pars{\alpha}
\end{align}

\begin{align}
I&=\ \overbrace{\int\root{\tan\pars{x} - \beta\over \tan\pars{x} + \beta}\,\dd x}
^{\ds{\mbox{Set}\ x \equiv t/2\ \imp\ t = 2x}}\ =\ 
\half\
\overbrace{\int\root{\tan\pars{t/2} - \beta\over \tan\pars{t/2} + \beta}\,\dd t}
^{\ds{\mbox{Set}\ y \equiv \tan\pars{t/2}\ \imp\ t = 2\arctan\pars{y}}}
\\[3mm]&=\half\int\root{y - \beta \over y + \beta}\,{2\,\dd y \over 1 + y^{2}}
\end{align}

With $\ds{{y - \beta \over y + \beta} \equiv z}$:
\begin{align}
I&=2\beta\int
{z\,\dd z \over \pars{\beta^{2} + 1}z^{2} + 2\pars{\beta^{2} - 1}z + \beta^{2} + 1}
\\[3mm]&={2\beta \over \beta^{2} + 1}\int
{z\,\dd z \over z^{2} + 2\bracks{\pars{\beta^{2} - 1}/\pars{\beta^{2} + 1}}z + 1}
=\sin\pars{2\alpha}
\int
{z\,\dd z \over z^{2} - 2\cos\pars{2\alpha}z + 1}
\end{align}
You can take it from here.
A: Given that,
$$I = \int \sqrt{ \dfrac {\sin(x-\alpha)} {\sin(x+\alpha)} }\,dx$$
multiplying and dividing by $\sqrt{\sin(x-\alpha)}$.
we get,
$$I = \int { \dfrac {\sin(x-\alpha)} {\sqrt{\sin(x+\alpha)\cdot \sin (x-\alpha) }}}\,dx$$
$$I=\int \dfrac{ \sin x\cdot \cos{\alpha }}{\sqrt{\sin^2x-\sin^2\alpha}}\ dx-\int\dfrac{\cos x\cdot \sin\alpha} {\sqrt{\sin^2x-\sin^2\alpha}}\ dx$$(how?)
$$I= \cos{\alpha}\int\dfrac{ \sin x dx}{\sqrt{\sin^2x-\sin^2\alpha}}dx-\sin\alpha\int\dfrac{\cos x dx}{\sqrt{\sin^2x-\sin^2\alpha}}dx$$
For the first integral,make the substituion $\cos x=u$.  For the second integral make the substituion $\sin x=v$.
You can take it from here.
