AP EAMCET · Maths · Indefinite Integration
\(\int \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x\) is equal to
- A \(\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+c\)
- B \(\frac{1}{2}\left(x \cos ^{-1} x+\sqrt{1-x^2}\right)+c\)
- C \(\frac{1}{2}\left(x \sin ^{-1} x-\sqrt{1-x^2}\right)+c\)
- D \(\frac{1}{2}\left(x \sin ^{-1} x+\sqrt{1-x^2}\right)+c\)
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 \(x=\cos 2 \theta\)…
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