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By using the properties of definite integrals, evaluate the integrals
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
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Let $\mathrm{I}=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \quad \ldots(i)$
Then $\mathrm{I}=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \sin x} d x \quad \ldots(ii)$
Adding (i) and (ii), we get
$\begin{aligned}
&2 \mathrm{I}=\int_0^{\pi / 2} \frac{\sin x-\cos x+\cos x-\sin x}{1+\sin x \cos x} d x \\
&=\int_0^{\pi / 2} 0 d x=0 \Rightarrow \mathrm{I}=0
\end{aligned}$
Then $\mathrm{I}=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \sin x} d x \quad \ldots(ii)$
Adding (i) and (ii), we get
$\begin{aligned}
&2 \mathrm{I}=\int_0^{\pi / 2} \frac{\sin x-\cos x+\cos x-\sin x}{1+\sin x \cos x} d x \\
&=\int_0^{\pi / 2} 0 d x=0 \Rightarrow \mathrm{I}=0
\end{aligned}$
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