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TS EAMCET · Maths · Indefinite Integration

\(\int \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x\) is equal to

  1. A \(\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+c\)
  2. B \(\frac{1}{2}\left(x \cos ^{-1} x+\sqrt{1-x^2}\right)+c\)
  3. C \(\frac{1}{2}\left(x \sin ^{-1} x-\sqrt{1-x^2}\right)+c\)
  4. D \(\frac{1}{2}\left(x \sin ^{-1} x+\sqrt{1-x^2}\right)+c\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+c\)

Step-by-step Solution

Detailed explanation

Let \(I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x\) Put \(x=\cos 2 \theta\)…