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$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ is equal to
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Let $I=\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cos x} d x$
On putting $x=\left(\frac{\pi}{2}-x\right)$ in Eq. (i), we get
$$
\begin{aligned}
I &=\int_{0}^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1-\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
&=\int_{0}^{\pi / 2} \frac{\cos x-\sin x}{1-\sin x \cos x} d x \\
&=-\int_{0}^{\pi / 2}\left(\frac{\sin x-\cos x}{1-\sin x \cos x}\right) d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\int_{0}^{\pi / 2} 0 d x=0
$$
$\Rightarrow \quad I=0$
On putting $x=\left(\frac{\pi}{2}-x\right)$ in Eq. (i), we get
$$
\begin{aligned}
I &=\int_{0}^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1-\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
&=\int_{0}^{\pi / 2} \frac{\cos x-\sin x}{1-\sin x \cos x} d x \\
&=-\int_{0}^{\pi / 2}\left(\frac{\sin x-\cos x}{1-\sin x \cos x}\right) d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\int_{0}^{\pi / 2} 0 d x=0
$$
$\Rightarrow \quad I=0$
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