TITLE: Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
QUESTION [21 upvotes]: 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:

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

REPLY [6 votes]: 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
$$
\cos(a)\cos^{-1}\left(\frac{\cos(x)}{\cos(a)}\right)-\sin(a)\log\left(\sin(x)+\sqrt{\sin^2(x)-\sin^2(a)}\right)+C'\tag{3}
$$