1
$\begingroup$

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.

$\endgroup$

closed as off-topic by Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500 Aug 28 '18 at 9:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Theoretical Economist, Leucippus, Jendrik Stelzner, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ Is this what you meant: $$\int \frac{\sqrt{\cos 2q}}{\sin q} \, \mathrm{d}q$$ $\endgroup$ – Tolaso Aug 27 '18 at 12:30
  • 1
    $\begingroup$ 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. $\endgroup$ – José Carlos Santos Aug 27 '18 at 12:31
  • 1
    $\begingroup$ Wolfram Alpha looks rather unpleasant $\endgroup$ – gt6989b Aug 27 '18 at 12:36
  • $\begingroup$ Tip: expand the double-angle cosine using the double angle formula. $\endgroup$ – FGSUZ Aug 27 '18 at 13:05
  • $\begingroup$ I done that still can't solve $\endgroup$ – Rajesh Gupta Aug 27 '18 at 13:09
1
$\begingroup$

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.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.