MCQ
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ is equal to :
  • A
    $\pi$
  • Zero
  • C
    $\int_0^{\pi / 2} \frac{2 \sin x}{1+\sin x \cos x} d x$
  • D
    $\frac{\pi^2}{4}$

Answer

Correct option: B.
Zero
Let $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\sin (\pi / 2-x)-\cos (\pi / 2-x)}{1+\sin (\pi / 2-x) \cos (\pi / 2-x)} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \cdot \sin x} d x$
Adding $(i)$ and $(ii)$, we get
$2 I=\int_0^{\pi / 2} 0 d x=0$
$\Rightarrow I=0$

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