# What is $\int \sqrt{\cos(2Q)} / {\sin(Q)} \,\mathrm{d}Q$? [closed]

What is $$\int \frac{\sqrt{\cos(2Q)}}{\sin(Q)} \,\mathrm{d}Q?$$

I have tried all the method which is possible but could not able to find the solution. Can anyone please tell me the solution of this problem.

## closed as off-topic by Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500Aug 28 '18 at 9:12

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• Is this what you meant: $$\int \frac{\sqrt{\cos 2q}}{\sin q} \, \mathrm{d}q$$ – Tolaso Aug 27 '18 at 12:30
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 27 '18 at 12:31
• Wolfram Alpha looks rather unpleasant – gt6989b Aug 27 '18 at 12:36
• Tip: expand the double-angle cosine using the double angle formula. – FGSUZ Aug 27 '18 at 13:05
• I done that still can't solve – Rajesh Gupta Aug 27 '18 at 13:09

## 1 Answer

First notice that $$\cos\left(2x\right)=\cos^2\left(x\right)-\sin^2\left(x\right)$$ and $$\sin\left(x\right)=\dfrac{\tan\left(x\right)}{\sec\left(x\right)}$$ so $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x ={\displaystyle\int} \sec^2(x) \cdot{{\dfrac{\sqrt{1-\tan^2\left(x\right)}}{\tan\left(x\right)\left(\tan^2\left(x\right)+1\right)}}}\,\mathrm{d}x$$ Change of variable as $$u = \tan (x)$$ we get $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x ={\displaystyle\int}\dfrac{\sqrt{1-u^2}}{u\left(u^2+1\right)}\,\mathrm{d}u$$ Another change of variable as $$v = \sqrt{1 - u^2}$$ gives us $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x =-{\displaystyle\int}\dfrac{v^2}{\left(v^2-2\right)\left(v^2-1\right)}\,\mathrm{d}v$$ Now let's factor the denominator as $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x ={\displaystyle\int}\dfrac{v^2}{\left(v-1\right)\left(v+1\right)\left(v^2-2\right)}\,\mathrm{d}v$$ Then perform partial fraction decomposition $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x ={\displaystyle\int}\left(\dfrac{2}{v^2-2}+\dfrac{1}{2\left(v+1\right)}-\dfrac{1}{2\left(v-1\right)}\right)\mathrm{d}v = 2A + \frac{1}{2}B - \frac{1}{2} C$$ Let's do $A$, $$A = {\displaystyle\int}\dfrac{1}{v^2-2}\,\mathrm{d}v = ={\displaystyle\int}\dfrac{1}{\left(v-\sqrt{2}\right)\left(v+\sqrt{2}\right)}\,\mathrm{d}v ={\displaystyle\int}\left(\dfrac{1}{2^\frac{3}{2}\left(v-\sqrt{2}\right)}-\dfrac{1}{2^\frac{3}{2}\left(v+\sqrt{2}\right)}\right)\mathrm{d}v$$ which s $$A = =\dfrac{\ln\left(v-\sqrt{2}\right)}{2^\frac{3}{2}}-\dfrac{\ln\left(v+\sqrt{2}\right)}{2^\frac{3}{2}}$$ Now, similarly $$B =\ln\left(v+1\right)$$ and $$C =\ln\left(v-1\right)$$ Plugging all $A,B,C$ back we get $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x =-\dfrac{\ln\left(v+\sqrt{2}\right)}{\sqrt{2}}+\dfrac{\ln\left(v-\sqrt{2}\right)}{\sqrt{2}}+\dfrac{\ln\left(v+1\right)}{2}-\dfrac{\ln\left(v-1\right)}{2}$$ Undoing the change of variable $v = \sqrt{1 - u^2}$, we get $${\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x=\dfrac{\ln\left(\sqrt{1-u^2}+\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-u^2}-\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-u^2}+1\right)}{2}+\dfrac{\ln\left(\sqrt{1-u^2}-1\right)}{2}$$ Undoing the other change of variable $u = \tan (x)$, we get $$\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}+\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}-\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}+1\right)}{2}+\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}-1\right)}{2}$$ Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.

• what is the mistake @RajeshGupta? – Ahmad Bazzi Aug 27 '18 at 15:51