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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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    Doesn't the inside reduce to $1/(1+t)^2$?2012-11-23
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    yes, but what then?2012-11-23
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    Beside to Issac's comment, can't you solve $\int (1/u^2)du?$2012-11-23
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    It would be -1/u + c. So... 1/(1+t)^2 should be -1/(1+t) + c... after substitution -1/(1+cos x) + c, but right answer is 1/(1+cos x)... then why?2012-11-23
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    Check the sign. Using your substitution $t=\cos x$ you should get $$\int \frac{1-\cos x}{\left( 1+\cos x\right) \sin x}dx=-\int \frac{1-t}{ \left( 1+t\right) \left( 1-t^{2}\right) }dt=-\int \frac{1}{\left( 1+t\right) ^{2}}dt$$2012-11-23
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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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    good idea, thanks :)2012-11-23
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    @user50222: Welcome; but I just wrote what Issacs note. :)2012-11-23
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    You are so gracious and humble! +12013-04-07