Question

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

Answer 1

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

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

$=-\int \frac{dx}{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}-1}$

$=-\int \frac{dx}{\frac{1-{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}-1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}$

$=-\int \frac{1+{\tan }^2\frac{x}{2}dx}{-2{\tan }^2\frac{x}{2}+2{\tan }^2\frac{x}{2}}$

Let $\tan \frac{x}{2}=u$ $\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=du$

So, by replacing the values we get

$I=\frac{1}{2}\int \frac{{\sec }^2\frac{x}{2}dx}{\tan \frac{x^2}{2}-\tan \frac{x}{2}}$

$=\int \frac{du}{u^2-u}$

$=\int \frac{1}{(1-\frac{1}{u})u^2}du$

Let $t=1-\frac{1}{u},\frac{dt}{du}=\frac{1}{u^2}$

$I=-\int \frac{1}{t}dt$

$=\ln \left(t\right)+c$

$=\ln (1-\frac{1}{u})+c$

$=\ln (1-\frac{1}{\tan \frac{x}{2}})+c$