$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$
$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$
0
$\frac{\pi}{2}$
1
$\frac{\pi}{4}$
Solution
$\text { Let } \begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x \\
I &=\int_{0}^{\frac{\pi}{2}} \frac{1-\frac{\cos x}{\sin x}}{\frac{1}{\sin x}+\cos x} d x=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x ...(1)\\
I &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{x}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
I &=\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x} d x ...(2)
\end{aligned}$
Adding equations (1) \& (2), we get
$\begin{array}{l}
2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x+\cos x-\sin x}{1+\sin x \cos x} d x \\
2 I=0 \Rightarrow I=0
\end{array}$