# Trigonometric Anti-derivative

What is $$\int \frac{\sin(x)^2}{\cos(x) + 1}dx\;?$$ I've tried everything I can think of, but I can't get it into a form that I can solve.

Note that $$\frac{\sin^2x}{1+\cos x}=\frac{1+\cos x}{1+\cos x}(1-\cos x)$$ since $1-\cos^2x=\sin ^2x$ and $1-y^2=(1-y)(1+y)$
• Why not: HINT: $\sin^2x=1-\cos^x=\ldots\;$? Oh, well. :-) (+1) – Brian M. Scott Jun 29 '13 at 1:29
• Argghhh. That should of course have been $1-\cos^2x$; the $2$ seems to have gone AWOL. – Brian M. Scott Jun 29 '13 at 1:42
• @BrianM.Scott That happened to kahen yesterday but with $\log^2x$. Curious. – Pedro Tamaroff Jun 29 '13 at 1:47