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Question:

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$

Solution:

Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$                ...(1)

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$                               $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x} d x$            ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\frac{\pi}{2}} \frac{0}{1+\sin x \cos x} d x$

$\Rightarrow I=0$

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