Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to compute $$ \int \frac{\sin x}{2\cos 2x} \operatorname{d}x$$

share|cite|improve this question
In an exam that would be -5 points for missing $\operatorname{d}x$... :-) – Asaf Karagila Oct 24 '11 at 8:02
up vote 2 down vote accepted

Note that $\frac{\sin x}{\cos2x}=\frac{1}{2}\frac{(\cos x+\sin x)-(\cos x-\sin x)}{(\cos x+\sin x)(\cos x-\sin x)}=\frac{1}{2}\left[\frac{1}{(\cos x-\sin x)}-\frac{1}{(\cos x+\sin x)}\right]=\frac{1}{2\sqrt{2}}[$sec$(x$-$\frac{\pi}{4})$-sec$(x$+$\frac{\pi}{4})]$

share|cite|improve this answer

Substituting $\cos x = y$, the integral becomes $$ \frac{1}{2} \int \frac{(-dy)}{2y^2-1} = \frac{1}{4} \int \frac{1}{\frac{1}{2}-y^2} dy. $$ Since the integrand is a rational function of $y$, we can write it in terms of partial fractions: $$ \frac{1}{\frac{1}{2}-y^2} = \frac{1}{\sqrt{2}} \left(\frac{1}{\frac{1}{\sqrt 2} +y} + \frac{1}{\frac{1}{\sqrt 2}-y} \right). $$ Can you take it from here?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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