# Where did I mistake to integrate $I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx\; ?$

It was given to integrate $$I=\int\sqrt{\frac{\sin(x-\alpha)}{\sin(x+\alpha)}} \; dx.$$

# Attempt:

\begin{align}I&= \int\sqrt{\frac{\sin((x+\alpha)-2\alpha)}{\sin(x+\alpha)}}\; dx\\&= \sqrt{\sin 2\alpha}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\;dx\; .\end{align}

Then I took \begin{align}\cot 2\alpha - \cot (x+\alpha)= z^2 \\ \implies [1 + \cot^2(x+\alpha)]\;dx= 2z\;dz\\ \implies [1+\{\cot 2\alpha - z^2\}^2]\; dx= 2z\;dz\; .\end{align}

Now, \begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int\frac{z\cdot 2z}{1+\{\cot 2\alpha - z^2\}^2}\;dz\\ &= \int\frac{2}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz \; .\end{align}

This makes me to break the integral to get \begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ \\ &= \int\frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z-\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right) +\text{const.}}\; dz +\int\frac{\left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z+\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right) +\text{const.}}\; dz \\ &=-\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}-\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}} .\end{align}

So, $$I= -\sqrt{\sin 2\alpha}\left(\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}+\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}}\right)\; .$$

Unfortunately, my book gave the answer as $$I= \cos\alpha \cos^{-1} (\cos x\sec\alpha) - \sin\alpha\log\left(\sin x+ \sqrt{\sin^2 x - \sin^2\alpha}\right)\; .$$

This is not near to my solution; I've made somewhere a big plunder.

So, could anyone please tell me where I did mistake? Is my attempted technique wrong?

• your mistake is in this step: \begin{align}I&= \int\sqrt{\frac{\sin((x+\alpha)-2\alpha)}{\sin(x+\alpha)}}\; dx\\&= \sqrt{\csc 2\alpha}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\;dx\; .\end{align} – Ekaveera Kumar Sharma Nov 14 '15 at 17:01
• @Ekaveera Kumar Sharma: Oh! sorry sir. I've changed it. But this also doesn't make the correct answer:( – user142971 Nov 14 '15 at 18:13
• your technique is absolutely ok. but can you explain this step: This makes me to break the integral to get $$-\frac{1}{z- \dfrac{(1+ \cot^2 2\alpha)}{z}}-\frac{1}{z+ \dfrac{(1+ \cot^2 2\alpha)}{z}} .$$ – Ekaveera Kumar Sharma Nov 15 '15 at 2:51
• @Ekaveera Kumar Sharma: Just, wait a minute, sir; I'll do that. – user142971 Nov 15 '15 at 2:52
• @Ekaveera Kumar Sharma: Check now if I could clear up. – user142971 Nov 15 '15 at 3:03

HINTS: We have $$\frac{\sin(x-\alpha)}{\sin(x+\alpha)}=\frac{\sin x\cos\alpha-\sin\alpha\cos x}{\sin x\cos\alpha+\sin\alpha\cos x}=\frac{(\sin x\cos\alpha-\sin\alpha\cos x)^2}{\sin^2x\cos^2\alpha-\sin^2\alpha\cos^2x}=\frac{(\sin x\cos\alpha-\sin\alpha\cos x)^2}{\sin^2x-\sin^2\alpha}$$

and $$u=(\sin^2x-\sin^2\alpha)^{1/2}\rightarrow du=\frac{\sin x\cos x}{\sqrt{\sin^2x-\sin^2\alpha}}dx$$

Can you go further?

• Thanks, for the answer, sir. But, my question was where I did the mistake & whether my way to integrate that was correct or not. I know this method but wanted to do it in other way. So, It would be great if you could help me point out where I mistook in my attempt. – user142971 Nov 15 '15 at 2:06

\begin{align}\int\sqrt{\cot 2\alpha - \cot (x+\alpha)}\; dx\\ &= \int \frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)+ \left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\dfrac{(1+ \cot^2 2\alpha)}{z^2} + z^2 -2\cot2\alpha}\; dz\\ \\ &= \int\frac{\left(1+ \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z-\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right)^2 +2\tan\alpha}\; dz +\int\frac{\left(1- \dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z^2}\right)}{\left(z+\dfrac{\sqrt{(1+ \cot^2 2\alpha)}}{z}\right)^2 -2\cot\alpha}\; dz \\ &=\int\frac{du}{u^2+2\tan\alpha}+\int\frac{dv}{v^2-2\cot\alpha}\end{align}

where $$u=z-\frac{\sqrt{1+\cot^2 2\alpha}}{z}$$ and $$v=z+\frac{\sqrt{1+\cot^2 2\alpha}}{z}$$

• We do have $log$ by this. The second integral with v gives $\frac1{2\sqrt{2\cot\alpha}}\log|\frac{v-\sqrt{2\cot\alpha}}{v+\sqrt2\cot\alpha}|$ . I think it is the same thing to $- \sin\alpha\log\left(\sin x+ \sqrt{\sin^2 x - \sin^2\alpha}\right)\;$, but the simplification is horrible. – joefu Nov 15 '15 at 6:47
• @joefu: Oh! now I could understand where I mistook. I used the wrong formula:( Thanks for the point. However, I've taken a lengthy procedure to solve that. I wonder how math professionals know which process to integrate is easier:/ – user142971 Nov 15 '15 at 7:55