Q53 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
$$
\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=
$$
- A$\frac{\pi}{4}$
- B$\frac{2}{\pi}$
- Correct answerC0
- D$\frac{\pi}{2}$
Solution
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
$$
2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 0 \mathrm{dx} \Rightarrow \mathrm{I}=0
$$
Hence, the correct answer is option: 'C'.