int {frac{{dx}}{{1 - cos x - sin x}}} = | vedclass

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(C) Let $I = \int \frac{dx}{1 - \cos x - \sin x}$.Using half-angle formulas $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ and $\sin x = \frac{2

Vedclass · Accepted Answer

$\log |1 - \cot(x/2)| + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{1 - \cos x - \sin x}$.Using half-angle formulas $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$ and $\sin x = \frac{2	an(x/2)}{1 + 	an^2(x/2)}$,we get:$I = \int \frac{dx}{1 - \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} - \frac{2	an(x/2)}{1 + 	an^2(x/2)}}$$I = \int \frac{(1 + 	an^2(x/2)) dx}{1 + 	an^2(x/2) - 1 + 	an^2(x/2) - 2	an(x/2)}$$I = \int \frac{\sec^2(x/2) dx}{2	an^2(x/2) - 2	an(x/2)}$Let $t = 	an(x/2)$,then $dt = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2 dt$.$I = \int \frac{2 dt}{2(t^2 - t)} = \int \frac{dt}{t(t - 1)}$Using partial fractions: $\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac{1}{t}$.$I = \int \left( \frac{1}{t - 1} - \frac{1}{t} ight) dt = \log|t - 1| - \log|t| + c = \log\left| \frac{t - 1}{t} ight| + c$$I = \log\left| \frac{	an(x/2) - 1}{	an(x/2)} ight| + c = \log|1 - \cot(x/2)| + c$.

$\int {\frac{{dx}}{{1 - \cos x - \sin x}}} = $

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