Question

If $\int \frac{1}{1-\cos x-\sin x}dx=\ln \left|\frac{\tan \left[f\right(x\left)\right]-1}{\tan \left[g\right(x\left)\right]}\right|+C,$ then the value of $f\left(1\right)+g\left(1\right)-1=$
(where $C$ is integration constant)

Answer 1

$I=\int \frac{1}{1-\cos x-\sin x}dx\Rightarrow I=\int \frac{1+{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}-(1-{\tan }^2\frac{x}{2})-2\tan \frac{x}{2}}dx[\because \sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}},\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}]\Rightarrow I=\int \frac{{\sec }^2\frac{x}{2}}{2{\tan }^2\frac{x}{2}-2\tan \frac{x}{2}}dx$
Put $\tan \frac{x}{2}=t$
$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt\Rightarrow I=\int \frac{dt}{t^2-t}\Rightarrow I=\int \frac{dt}{t(t-1)}\Rightarrow I=\int \frac{dt}{t-1}-\int \frac{dt}{t}\Rightarrow I=\ln |t-1|-\ln |t|+C\Rightarrow I=\ln \left|\frac{t-1}{t}\right|+C\Rightarrow I=\ln \left|\frac{\tan \frac{x}{2}-1}{\tan \frac{x}{2}}\right|+C\Rightarrow f\left(x\right)=\frac{x}{2}=g\left(x\right)$
So,
$f\left(1\right)+g\left(1\right)-1=0$