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MHT CET · Maths · Definite Integration

\(\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x\) is equal to

  1. A 0
  2. B \(\frac{\pi}{2}\)
  3. C \(\frac{\pi}{4}\)
  4. D \(\pi\)
Verified Solution

Answer & Solution

Correct Answer

(A) 0

Step-by-step Solution

Detailed explanation

Let \(I=\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cos x} d x\)
On putting \(x=\left(\frac{\pi}{2}-x\right)\) in Eq. (i), we get
\(
\begin{aligned}
I &=\int_{0}^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1-\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
&=\int_{0}^{\pi / 2} \frac{\cos x-\sin x}{1-\sin x \cos x} d x \\
&=-\int_{0}^{\pi / 2}\left(\frac{\sin x-\cos x}{1-\sin x \cos x}\right) d x
\end{aligned}
\)
On adding Eqs. (i) and (ii), we get
\(
2 I=\int_{0}^{\pi / 2} 0 d x=0
\)
\(\Rightarrow \quad I=0\)