Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$ 
Integrate   $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$


Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}dx\end{align}$$
Now: 
$$\begin{align}u^2&=\frac{\sin(x+a-2a)}{\sin(x+a)}
\\u^2&=\frac{\sin(x+a)\cos(2a)-\cos(x+a)\sin(2a)}{\sin(x+a)}
\\u^2&=\cos(2a)-\sin(2a)\cot(x+a)
\\\cot(x+a)&=(\cos(2a)-u^2)\csc(2a)
\\\csc^2(x+a)=\cot^2(x+a)+1&=(\cos(2a)-u^2)^2\csc^2(2a)+1
\\\csc^2(x+a)&=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}
\\\sin^2(x+a)&=\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}\end{align}$$
Now:
$$\begin{align}
I&=\int u.\frac{2udu\sin^2(x+a)}{\sin(2a)}\\
I&=\int\frac2{\sin(2a)}.u^2.\frac{\sin^2(2a)}{u^4-2u^2\cos(2a)+1}du
\\I&=2\sin(2a)\int\frac{u^2}{u^4-2ku^2+1}du\quad k:=\cos 2a
\\\frac If&=\int\frac{2+u^{-2}-u^{-2}}{u^2-2k+u^{-2}}du=\int\frac{1+u^{-2}}{u^2-2k+u^{-2}}du+\int\frac{1-u^{-2}}{u^2-2k+u^{-2}}du\quad \\f:=\sin(2a)
\\&=\int\frac{d(u-u^{-1})}{(u-u^{-1})^2+2-2k}+\int\frac{d(u+u^{-1})}{(u+u^{-1})^2-2-2k}\end{align}$$
Now: $2-2k=2(1-\cos 2a)=4\sin^24a,2+2k=2(1+\cos 2a)=4\cos^24a$
So:
$$I=\sin2a\left(\frac1{2\sin4a}\arctan\left(\frac{u-u^{-1}}{2\sin(4a)}\right)+\frac1{4\cos4a}\ln\left|\frac{u+u^{-1}-2\cos 4a}{u+u^{-1}+2\cos 4a}\right|\right)+C$$
Or:
$$I=\frac1{4\cos2a}\arctan\left(\frac{-\sin a\cos x}{\sin4a\sqrt{\sin(x+a)\sin(x-a)}}\right)+\frac{\sin2a}{4\cos 4a}\ln\left|\frac{\sin x\cos a-\cos4a\sqrt{\sin(x+a)\sin(x-a)}}{\sin x\cos a+\cos4a\sqrt{\sin(x+a)\sin(x-a)}}\right|+C$$
But the textbook answer is:

$$\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+c$$

 A: Too long for a comment, perhaps the following approach would help.
Express the inner term of square root as follows
\begin{align}
\frac{\sin(x-a)}{\sin(x+a)}&=\frac{\sin(x)\cos(a)-\cos(x)\sin(a)}{\sin(x)\cos(a)+\cos(x)\sin(a)}\qquad\Rightarrow\qquad\text{divide by}\,\cos(x)\cos(a)\\
&=\frac{\tan(x)-\tan(a)}{\tan(x)+\tan(a)}
\end{align}
Let $t^2=\tan(x)+\tan(a)$, then
\begin{align}
\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\,dx&=\int\sqrt{\frac{\tan(x)-\tan(a)}{\tan(x)+\tan(a)}}\,dx\\
&=2\int\frac{\sqrt{t^2-2\tan(a)}}{1+(t^2-\tan(a))^2}\,dt\\
&=2\int\frac{\sqrt{1-2\frac{\tan(a)}{t^2}}}{\frac{1}{t^4}+\left(1-\frac{\tan(a)}{t^2}\right)^2}\,\frac{dt}{t^3}\qquad\Rightarrow\qquad\text{let}\,u=\frac{\tan(a)}{t^2}\\
&=-\int\frac{\sqrt{1-2u}}{\frac{u^2}{\tan^2(a)}+\left(1-u\right)^2}\,\frac{du}{\tan(a)}\\
&=-\int\frac{\tan(a)\sqrt{1-2u}}{u^2\sec^2(a)-2u\tan^2(a)+\tan^2(a)}\,du\\
&=-\frac{1}{\tan(a)}\int\frac{\sqrt{1-2u}}{u^2\csc^2(a)-2u+1}\,du\\
\end{align}
A: Other answers offer alternative approaches to integrating form the beginning. If you are looking to find where your steps went astray, it starts when  you are using: $$2(1-\cos(2a))=4\sin^2(4a)$$ but the correct identity is: $$2(1-\cos(2a))=4\sin^2(a)$$ Just check both sides with $a=\pi/4$ and you'll believe it. And similarly for the next identity that you use: $2(1+\cos(2a))=4\cos^2(4a)$ should be $2(1+\cos(2a))=4\cos^2(a)$.
From there, you would have 
$$I=\sin2a\left(\frac1{2\sin a}\arctan\left(\frac{u-u^{-1}}{2\sin(a)}\right)+\frac1{4\cos a}\ln\left|\frac{u+u^{-1}-2\cos a}{u+u^{-1}+2\cos a}\right|\right)+C$$
which gives
$$I=\cos(a)\arctan\left(\frac{u-u^{-1}}{2\sin(a)}\right)+\frac1{2}\sin(a)\ln\left|\frac{u+u^{-1}-2\cos a}{u+u^{-1}+2\cos a}\right|+C$$
And then we start back-substituting. We reach a  point where it helps to use $\arctan\left(\frac{A}{B}\right)=\arcsin\left(\frac{A}{\sqrt{A^2+B^2}}\right)=\frac{\pi}{2}-\arccos\left(\frac{A}{\sqrt{A^2+B^2}}\right)$ and the constant can be absorbed into the $C$. Keep in mind that expressions in $a$ are constant for the purposes of $C$. Then we reach a point where by adding a certain logarithmic expression in $a$ (again, absorbed into $C$) we can simplify the appearance of the logarithmic term.
$$
\begin{align}
I&=\cos(a)\arctan\left(\frac{\sqrt\frac{\sin(x-a)}{\sin(x+a)}-\sqrt\frac{\sin(x+a)}{\sin(x-a)}}{2\sin(a)}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sqrt\frac{\sin(x-a)}{\sin(x+a)}+\sqrt\frac{\sin(x+a)}{\sin(x-a)}-2\cos a}{\sqrt\frac{\sin(x-a)}{\sin(x+a)}+\sqrt\frac{\sin(x+a)}{\sin(x-a)}+2\cos a}\right|+C\\
&=\cos(a)\arctan\left(\frac{\sin(x-a)-\sin(x+a)}{2\sin(a)\sqrt{\sin(x-a)\sin(x+a)}}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x-a)+\sin(x+a)-2\cos(a)\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x-a)+\sin(x+a)+2\cos(a)\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=\cos(a)\arctan\left(\frac{-\cos(x)}{\sqrt{\sin(x-a)\sin(x+a)}}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=-\cos(a)\arcsin\left(\frac{\cos(x)}{\sqrt{\cos^2(x)+\sin(x-a)\sin(x+a)}}\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=-\cos(a)\left(\frac{\pi}{2}-\arccos\left(\frac{\cos(x)}{\sqrt{\cos^2(x)+\frac12\cos(2a)-\frac12\cos(2x)}}\right)\right)\\
&\phantom{=}{}+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C\\
&=\cos(a)\arccos\left(\frac{\cos(x)}{\cos(a)}\right)+\frac1{2}\sin(a)\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|+C_1
\end{align}$$
Note that $-\cos(a)\frac{\pi}{2}$ has been absorbed into $C$.
It remains to show $$\frac1{2}\ln\left|\frac{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|=-\ln\left|\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right|+C_2(a)$$
If you add $\ln\left|\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right|$ to the left side, you have $$
\begin{align}
&\ln\left|\sqrt{\sin(x)-\sqrt{\sin(x-a)\sin(x+a)}}\sqrt{\sin(x)+\sqrt{\sin(x-a)\sin(x+a)}}\right|\\
&=\ln\left|\sqrt{\sin^2(x)-\sin(x-a)\sin(x+a)}\right|\\
&=\ln\left|\sqrt{\sin^2(x)-\frac12\cos(2a)+\frac12\cos(2x)}\right|\\
&=\ln\left|\sqrt{\sin^2(a)}\right|=C_2(a)\\
\end{align}$$
which establishes that last relation.

Also this write up has a typo at an earlier line, with $$\csc^2(x+a)=\frac{\cos^2(2a)+u^4-2u^2\cos2(2a)+\sin^2(2a)}{\csc^2(2a)}$$ but you meant $$\csc^2(x+a)=\frac{\cos^2(2a)+u^4-2u^2\cos(2a)+\sin^2(2a)}{\sin^2(2a)}$$ In any case, these things have been corrected at the next line.
A: Since $\sin(x-a)\sin(x+a)=\cos^2(a)-\cos^2(x)=\sin^2(x)-\sin^2(a)$, if we let
$$
u=\frac{\cos(x)}{\cos(a)}\quad\text{and}\quad v=\frac{\sin(x)}{\sin(a)}\tag{1}
$$
then
$$
\begin{align}
&\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\,\mathrm{d}x\\
&=\int\frac{\sin(x-a)}{\sqrt{\sin(x+a)\sin(x-a)}}\,\mathrm{d}x\\
&=\int\left[\frac{\sin(x)\cos(a)}{\sqrt{\cos^2(a)-\cos^2(x)}}-\frac{\cos(x)\sin(a)}{\sqrt{\sin^2(x)-\sin^2(a)}}\right]\,\mathrm{d}x\\
&=-\cos(a)\int\frac{\mathrm{d}u}{\sqrt{1-u^2}}-\sin(a)\int\frac{\mathrm{d}v}{\sqrt{v^2-1}}\\[6pt]
&=\cos(a)\cos^{-1}\left(\frac{\cos(x)}{\cos(a)}\right)-\sin(a)\cosh^{-1}\left(\frac{\sin(x)}{\sin(a)}\right)+C\tag{2}
\end{align}
$$
Since $\cosh^{-1}(x)=\log\left(x+\sqrt{x^2-1}\right)$, $(2)$ is equal to
$$
\cos(a)\cos^{-1}\left(\frac{\cos(x)}{\cos(a)}\right)-\sin(a)\log\left(\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right)+C'\tag{3}
$$
A: Here is an alternative way.
\begin{align}\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}&=\frac{\sin(x-a)}{\sqrt{\sin(x-a)\sin(x+a)}}=\frac{\sin x\cos a-\cos x\sin a}{\sqrt{\frac12(\cos 2a-\cos 2x)}}\\&=\frac{\cos a\sin x}{\sqrt{\cos^2a-\cos^2x}}-\frac{\sin a\cos x}{\sqrt{\sin^2 x-\sin^2 a}}\end{align}
Now for the first one, substitute $\cos x=\cos t\cos a$. 
$$\int \frac{\cos a\sin x}{\sqrt{\cos^2a-\cos^2x}}dx=\int\cos a\,dt=t\cos a=\cos a\arccos\left(\frac{\cos x}{\cos a}\right)+C_1$$
The second one, substitute $\sin x=\cosh s\sin a$.
\begin{align}\int\frac{\sin a\cos x}{\sqrt{\sin^2 x-\sin^2 a}}dx&=\int\sin a\,ds=s\sin a=\sin a\cosh^{-1}\left(\frac{\sin x}{\sin a}\right)\\&=\sin a\ln\left(\frac{\sin x}{\sin a}+\sqrt{\frac{\sin^2x}{\sin^2a}-1}\right)+C_2\\&=\sin a\ln(\sin x+\sqrt{\sin^2x-\sin^2a})+C_3\end{align}
Therefore, $$\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}dx=\cos a\arccos\left(\frac{\cos x}{\cos a}\right)-\sin a\ln(\sin x+\sqrt{\sin^2 x-\sin^2 a})+C$$
A: Let us write $$I_a=\int\sqrt{\frac{\sin(x-a)}{\sin(x+a)}}\mathrm dx$$
and compute $I_+=I_a+I_{-a}$ and $I_-=I_a-I_{-a}$. We get
$$I_\pm=I_a\pm I_{-a}=\int\frac{\sin(x-a)\pm\sin(x+a)}{\sqrt{\sin(x+a)\sin(x-a)}}\mathrm dx.$$
Using trigonometric formulas, one obtains
$$I_+=2\cos a\int\frac{\sin x}{\sqrt{\cos^2a-\cos^2x}}\mathrm dx\\
  I_-=2\sin a\int\frac{\cos x}{\sqrt{\sin^2x-\sin^2a}}\mathrm dx.$$
Now just apply the adequate changes of variables
$$I_+=2\cos a \arccos\left(\frac{\cos x}{\cos a}\right)+c_+\\
  I_-=-2\sin a \ln\left(\sin x+\sqrt{\sin^2x-\sin^2a}\right)+c_-$$
and the result is $I_a=\frac12\left(I_++I_-\right)$.
