MCQ
$\int_0^{\pi /2} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}} \,dx = $
  • A
    $2$
  • B
    $ - 2$
  • $0$
  • D
    None of these

Answer

Correct option: C.
$0$
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, $2I = 0 \Rightarrow I = 0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free