MCQ
$\int_0^{\frac{\pi}{2}} \frac{\cos 2 x}{\cos x+\sin x} d x=$
  • A
    -1
  • B
    1
  • $0$
  • D
    2

Answer

Correct option: C.
$0$
(C)
$\int_0^{\frac{\pi}{2}} \frac{\cos 2 x}{\cos x+\sin x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x-\sin ^2 x}{\cos x+\sin x} d x$
$\begin{array}{l}=\int_0^{\frac{\pi}{2}}(\cos x-\sin x) d x \\ =[\sin x+\cos x]_0^{\pi / 2}=0\end{array}$

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