ExamBro
ExamBro
MHT CET · Maths · Definite Integration

If \(\int_{0}^{1} \tan ^{-1} x d x=p\), then the value of \(\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) d x\) is

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

Answer & Solution

Correct Answer

(B) \(\frac{\pi}{4}-p\)

Step-by-step Solution

Detailed explanation

\(\int_{0}^{1}\left(\frac{1-x}{1+x}\right.) d x\)
\(=\int_{0}^{1}\left[\tan ^{-1}(1)-\tan ^{-1}(x)\right] d x\)
\(=\int_{0}^{1} \frac{\pi}{4} d x-\int_{0}^{1} \tan ^{-1}(x) d x\)
\(=\frac{\pi}{4}-p\)