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

Correct Answer is : 12(xcos−1x−√1−x2)+c

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

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

\frac{1}{2} (x \sin ^{-1} x-\sqrt{1-x^2}) +c

\frac{1}{2} (x \sin ^{-1} x+\sqrt{1-x^2}) +c

EAMCET Engineering 2007 Solved Paper

Section: Mathematics

Question : 152 of 160

Marks: +1, -0

∫tan−1(√‌

1−x

1+x

)‌dx is equal to

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