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
- A \(\frac{\pi}{4}+p\)
- B \(\frac{\pi}{4}-p\)
- C \(1+p\)
- D \(1-p\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}-p\)
Step-by-step Solution
Detailed explanation
\(\int_{0}^{1}\left(\frac{1-x}{1+x}\right.) d x\)
\(=\int_{0}^{1}\left[\tan ^{-1}(1)-\tan ^{-1}(x)\right] d x\)
\(=\int_{0}^{1} \frac{\pi}{4} d x-\int_{0}^{1} \tan ^{-1}(x) d x\)
\(=\frac{\pi}{4}-p\)
\(=\int_{0}^{1}\left[\tan ^{-1}(1)-\tan ^{-1}(x)\right] d x\)
\(=\int_{0}^{1} \frac{\pi}{4} d x-\int_{0}^{1} \tan ^{-1}(x) d x\)
\(=\frac{\pi}{4}-p\)
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