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

\[\begin {array} {1 1} (a)\;\frac{1}{2}(x \cos ^{-1} x -\sqrt {1-x^2})+c & \quad (b)\;\frac{1}{2}(x \cos ^{-1} x +\sqrt {1-x^2})+c \\ (c)\;\frac{1}{2}(x \sin ^{-1} x -\sqrt {1-x^2})+c & \quad (d)\;\frac{1}{2}(x \sin ^{-1} x +\sqrt {1-x^2})+c \end {array}\]

1 Answer

$(a)\;\frac{1}{2}(x \cos ^{-1} x -\sqrt {1-x^2})+c$
answered Nov 7, 2013 by pady_1
 
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