3
$\begingroup$

The integral is $$\int\frac{\sqrt{\cos 2x}}{\sin x}\,\text{d}x.$$ I have tried solving this by taking the sine inside the radical as follows: $$\int\sqrt\frac{\cos 2x}{\sin^2 x}\,\text{d}x$$$$\int\sqrt\frac{\cos^2x-\sin^2x}{\sin^2 x}\,\text{d}x$$$$\int\sqrt{\cot^2x-1}\,\text{d}x.$$ I don't know how to proceed from here, or whether this is even right. Any suggestions?

$\endgroup$
4
$\begingroup$

You can continue like this: \begin{align*} \int\sqrt{\cot^2(x)-1}\,dx&=\int\frac{\cot^2(x)-1}{\sqrt{\cot^2(x)-1}}\,dx\\ &=\int\frac{\csc^2(x)-2}{\sqrt{\cot^2(x)-1}}\,dx\\ &=\int\frac{\csc^2(x)}{\sqrt{\cot^2(x)-1}}\,dx-2\int\frac{1}{\sqrt{\cot^2(x)-1}}\,dx\\ \end{align*} The first integral can be solved with substitution $u=\cot(x)$ and the second one with substitution $\cot(x)=\cosh(t)$ will be converted to the following one: \begin{align*} \int\frac{1}{\sqrt{\cot^2(x)-1}}\,dx&=-\int\frac{dt}{1+\cosh^2(t)}\\ &=-\int\frac{2}{2+e^t+e^{-t}}\,dt\\ &=-\int\frac{2e^t}{e^{2t}+2e^t+1}\,dt\\ &=-\int\frac{2e^t}{(e^t+1)^2}\,dt\\ &=\frac{2}{e^t+1} \end{align*} Can you continue?( I hope you can)

$\endgroup$
  • $\begingroup$ Perhaps you might want to tackle this integral as well ? $\endgroup$ – Lucian Dec 14 '13 at 8:04
  • $\begingroup$ Yes I can :) Thank you :) $\endgroup$ – Artemisia Dec 15 '13 at 4:42
  • $\begingroup$ Nice Solution Farshad Nahangi.... $\endgroup$ – juantheron Dec 15 '13 at 11:55
1
$\begingroup$

Suppose you express everything in terms of $\cos(x)$ and later make a change of variables such that $y = \cos(x)$. Remember that $\frac{dx}{\sin(x)}$ is also $\frac{-d(\cos(x)) }{1-\cos^2(x)}$. Does this help and can you continue from here ?

$\endgroup$
  • $\begingroup$ @freak_warrior. Thanks for editing. I am almost blind and, beside plain ASCII, everything is difficult to me. Cheers. $\endgroup$ – Claude Leibovici Dec 14 '13 at 6:40
  • $\begingroup$ haha no problem $\endgroup$ – freak_warrior Dec 14 '13 at 6:42
  • $\begingroup$ @ClaudeLeibovici : Your answer is right as well :) I just can accept only one answer. But I like your method :) Thank you $\endgroup$ – Artemisia Dec 15 '13 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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