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$ \int \frac {1-\cos x}{(1+\cos x)\sin x} dx$

I tried to expand the fraction by sin x and substitute t = cos x, so I got $ \int \frac{(1-t)}{(1+t)(1-t)(1+t)} dt$ here i could cancel out (1-t)... but what next? I don't know which formula should be used.

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    You are absolutely right. I overlooked it.2012-11-23

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Hint: $\int\frac{1-\cos x}{(1+\cos x)\sin x}dx=\int\frac{1-\cos^2 x}{(1+\cos x)^2\sin x}dx=\int\frac{\sin^2 x}{(1+\cos x)^2\sin x}dx=\int\frac{\sin x}{(1+\cos x)^2}dx$ Now take $t=\cos x$ as you noted before.

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    You are so gracious and humble! +12013-04-07