MCQ
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$ is equal to :
- A$\pi$
- BZero (0)
- C$\int_0^{\pi / 2} \frac{2 \sin x}{1+\sin x \cos x} d x$
- D$\frac{\pi^2}{4}$
✓
Answer
Let $I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$
\begin{array}{l}
\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
\end{array}
$
Adding (i) and (ii), we get
$
2 I=\int_0^{\pi / 2} 0 d x=0 \Rightarrow I=0
$
$
\begin{array}{l}
\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
\end{array}
$
Adding (i) and (ii), we get
$
2 I=\int_0^{\pi / 2} 0 d x=0 \Rightarrow I=0
$
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