Thread: integrate ((sin(x))^2)/(1+cos(x))

I'm stuck.. can't remember how to integrate this (I'm sure I used to know)
How do I evaluate
$\displaystyle \int \dfrac{sin^{2}(x)}{1 + cos(x)}~dx$
I tried to do that z substitution thingy where
$\displaystyle z = tan(\dfrac{1}{2}x), sin(x) = \frac{2z}{1+z^{2}}, cos(x) = \frac{1-z^{2}}{1+z^{2}}, dx = \frac{2dz}{1+z^{2}}$
therefore
$\displaystyle \int \frac{sin^{2}(x)}{1 + cos(x)}~dx$ became $\displaystyle \int \frac{4z^{2}}{(1+z^{2})^{2}}~dz$
but I also still can't manage to evaluate it...
pls help guide me and also if you can help verify that my z substitution is correct please.

That last is a candidate for "partial fractions": find A, B, C, and D so that $\displaystyle \frac{4z^2}{(1+ z^2)^2}= \frac{Az+ B}{1+ z^2}+ \frac{Cz+ D}{(1+ z^2)^2}$.

Mutiplying both sides by $\displaystyle (1+ z^2)^2$, $\displaystyle 4z^2= (Az+ B)(1+ z^2)+ Cz+ D= Az^3+ Bz^2+ (A+ C)z+ (B+ D)$ so we must have A= 0, B= 4, A+ C= C= 0, and B+ D= 4+ D= 0 so D= -4.

This integral is the same as $\displaystyle 4\int \frac{1}{1+ z^2}dz- 4\int \frac{1}{(1+ z^2)^2}dz$

What you're doing might be an over kill. Try $\displaystyle \sin^2x = 1 - \cos^2x = (1 - \cos x)(1 + \cos x)$.

Originally Posted by Jester

What you're doing might be an over kill. Try $\displaystyle \sin^2x = 1 - \cos^2x = (1 - \cos x)(1 + \cos x)$.

Now that you have pointed that out I agree that what I did is definitely overkill.... I knew somehow that there was an easy way to solve it but I couldn't think of it... thnx jest
And Halls yes thank you I don't know why I didn't recognize that the form was ripe for using partial fraction decomp...
Actually today I had been solving several hard problems for hours and you know sometimes you look at something really simple and for some reason your brain stops working.....
anyway thanks to all.