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.
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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 $1-y^2=(1-y)(1+y)$
Pedro
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Why not: HINT: $\sin^2x=1-\cos^x=\ldots;$? Oh, well. :-) (+1) – Brian M. Scott Jun 29 '13 at 01:29
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Argghhh. That should of course have been $1-\cos^2x$; the $2$ seems to have gone AWOL. – Brian M. Scott Jun 29 '13 at 01:42
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Thank you very much for your help. – GregoryComer Jun 29 '13 at 01:44
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@BrianM.Scott That happened to kahen yesterday but with $\log^2x$. Curious. – Pedro Jun 29 '13 at 01:47