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I have an expression $\int_0^\frac\pi2 \frac1{\sin x} dx$ = $\int_0^\frac\pi2 \frac{\sin x}{1-cos^2x} dx = |subs: t=cos x; dt = -sinx dx| = \int_1^0 \frac{-1}{1-t^2} dt = \int_0^1 \frac{\frac12}{1-t} + \frac{\frac12}{1+t} $ why do I have to take -1 out of (1-t) and do $\frac12[\ln\frac{1+t}{t-1}]_0^1 $ instead of $\frac12[\ln(1-t^2)]_0^1 $ ?

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    If you want to get $\frac{1}{\sin x+\ln x}$ you should type \frac{1}{\sin x+\ln x}2012-11-23

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If I understand correctly, then you ask why $\int\frac{1}{1-t}=-\ln(t-1)$ and not $\ln(1-t)$. This happens because if try to differentiate $\ln(1-t)$, you will get $\frac{-1}{1-t}$, since, in general $\ln(f)'=\frac{f'}{f}$.