Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

2 Answers 2

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|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|improve this answer

Your Answer

 
discard

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.