Integrate $\sin(2x)/(1 + \cos^2x)$ with $u$-substitution

Question

Here's how I started: $$\int\frac{\sin{(2x)}}{1+\cos^2 x} dx = -2\int\frac{-\sin x\cos x}{1+\cos^2 x} dx = -2\int \frac{u}{1+u^2}du $$

I know the answer ends up being this from here:

$$ -\ln\left(1+\cos^2 x\right) $$

But I don't understand where this u in the numerator goes, it seems like the answer should look like this:

$$-2\cdot\frac{1}{2}\cdot u^2\cdot\ln\left(1+u^2\right) = -\cos^2 x \cdot \ln\left(1+\cos^2 x\right) $$

Thanks!

re-check my answer. There was an ugly bug in it...and you were right : there are logarithms in this question! — DonAntonio, Jan 19 '14 at 20:51

DonAntonio · Answer 1 · 2014-01-19T20:50:52.580

6

No substitution at all needed. Just a little trigonometry and some basic integration rules:

$$(\cos^2x)'=-2\sin x\cos x\;\;,\;\;\sin 2x=2\sin x\cos x$$

so

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

and now you may want to use the general rule

$$\int\frac{f'(x)}{f(x)}dx=\log f(x)+C$$

edited Jan 19 '14 at 20:50

answered Jan 19 '14 at 20:43

DonAntonio

You're completely right, @DanielFischer. Good catch and thanks. – DonAntonio Jan 19 '14 at 20:50
2

Thanks! I like that your way of going about it. – Joshua Oliphant Jan 19 '14 at 20:55
I had not seen that rule before about ∫f'(x)/f(x)dx = log(f(x)) + c but I will remember it. – Joshua Oliphant Jan 19 '14 at 21:10
I think this (and many more like this, with derivatives) makes things easier, clearer and shorter – DonAntonio Jan 19 '14 at 21:13

score 0 · Answer 2 · answered Jan 19 '14 at 21:05

0

Make the substitution $u=1+\cos^2x$. Then $du= -2\sin x\cos x=-\sin 2x$, so the integral becomes

$$- \int \frac{du}{u} = -\ln(1+\cos^2x) + C. $$

answered Jan 19 '14 at 21:05

Avi Steiner

2 Answers2