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.
Best Answer
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.