_0^ / 2 x- x 1+ x x d x= - Vidyadip

MCQ

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

A
2
B
-2
✓
$0$
D
1

✓

Answer

Correct option: C.

$0$

(C)
$\text { Let } I=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\sin x \cos x} d x$ ...(i)
$\therefore \quad I =\int_0^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I =\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\cos x \sin x} d x$ ...(ii)
Adding (i) and (ii), we get
$2 I =0 \Rightarrow I =0$

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