Q131 / 200 · Manipal (MET) 2014 · Mathematics
$\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cdot \cos x} d x$ is equal to
- Correct answerA0
- B$\frac{\pi}{4}$
- C$\frac{\pi}{2}$
- D$\pi$
Solution
$\begin{aligned}
& \text { Let } I=\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x ...(i)\\
& \begin{aligned}
& \Rightarrow I=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right] }
\end{aligned} \\
& \quad=\int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \sin x} d x ...(ii)\\
& {\left[\because \sin \left(\frac{\pi}{2}-x\right)=\cos x \text { and } \cos \left(\frac{\pi}{2}-x\right)=\sin x\right]}
\end{aligned}$
On adding Eqs. (i) and (ii), we get $2 I=\int_0^{\pi / 2} \frac{0}{1+\sin x \cos x} d x=0 \Rightarrow I=0$
On adding Eqs. (i) and (ii), we get $2 I=\int_0^{\pi / 2} \frac{0}{1+\sin x \cos x} d x=0 \Rightarrow I=0$
Hence, the correct answer is option: 'A'.