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

Question

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.

Answer 1

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