Question
$\int_0^{\pi /2} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}} \,dx = $

Answer

c
(c) $\int_0^{\pi /2} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}dx = I} $....$(i)$

Now $I = \int_0^{\pi /2} {\frac{{\cos \left( {\frac{\pi }{2} - x} \right) - \sin \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}\,dx} $

$= \int_0^{\pi /2} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}\,\,dx} $.....$(ii)$

On adding, $2I = 0 \Rightarrow I = 0$.

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