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

The question is to find

$$\displaystyle \int \frac {\sin (2x)}{1+\cos^2x}.$$

Can anyone help me? I need all the steps, because I need to understand what to do. Thank you.

share|cite|improve this question
up vote 6 down vote accepted

You can use the identity: $$ \sin(2x)=2\sin(x)\cos(x). $$ Then use a $u$-substitution with $u=1+\cos^2(x)$.

share|cite|improve this answer
I'm too slow formatting integrals, I think! – amWhy Dec 5 '12 at 20:24
Wow, that definately clears up alot. Sorry about any bad formatting when i made the question. I am new to this site. – Jeremy Rowler Dec 5 '12 at 20:30

By the double angle formula, $\sin(2x) = 2 \sin(x)\cos(x)$

$$\int \frac{\sin(2x)}{1 + \cos^2(x)}dx = \int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)}dx$$

Let $u = 1 + \cos^2(x)$ $du = -2\sin(x)\cos(x) dx$

so...substituting, we get: $$\int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)}dx = \int -\frac{1}{u} du$$

Can you take it from here?

Integrate with respect to $u$, then "back" substitute $u = 1 + \cos^2(x)$ into the result.

share|cite|improve this answer

Note that $\sin(2x)=2\sin(x)\cos(x)$

Therefore the problem reduces to finding the integral: $\int \frac {2\sin(x)\cos(x)}{1+\cos^2(x)}dx=-\log(1+\cos^2(x))+C$

share|cite|improve this answer
A photo finish. But, you beat me by a second. +1 – Joe Johnson 126 Dec 5 '12 at 20:20

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.