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$$
\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=
$$
Options:
\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=
$$
Solution:
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Verified Answer
The correct answer is:
0
Let $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cos x} d x$
$$
\begin{aligned}
& \therefore 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}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
& =\int_0^{\pi / 2} \frac{\cos x-\sin x}{1-\cos x \sin x} d x
\end{aligned}
$$
Eq. (1) $+(2)$ gives
$$
2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 0 \mathrm{dx} \Rightarrow \mathrm{I}=0
$$
$$
\begin{aligned}
& \therefore 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}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
& =\int_0^{\pi / 2} \frac{\cos x-\sin x}{1-\cos x \sin x} d x
\end{aligned}
$$
Eq. (1) $+(2)$ gives
$$
2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 0 \mathrm{dx} \Rightarrow \mathrm{I}=0
$$
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