int frac{dx}{1-cos x-sin x} is equal to | vedclass

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

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$\log \left|1-\cot \frac{x}{2}\right|+c$

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(C) We have,$I = \int \frac{dx}{1-\cos x-\sin x}$.Using the 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)} = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{	an(x/2)(	an(x/2)-1)}$.Let $z = 	an(x/2)$,then $dz = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2dz$.$I = \int \frac{2dz}{2z(z-1)} = \int \frac{dz}{z(z-1)}$.Using partial fractions: $\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$.$I = \int \left(\frac{1}{z-1} - \frac{1}{z}ight) dz = \log|z-1| - \log|z| + C = \log\left|\frac{z-1}{z}ight| + C$.Substituting $z = 	an(x/2)$ back:$I = \log\left|\frac{	an(x/2)-1}{	an(x/2)}ight| + C = \log|1 - \cot(x/2)| + C$.

$\int \frac{dx}{1-\cos x-\sin x}$ is equal to

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