$\smallint \tan \,({\sin ^{ - 1}}\,x)\,\;dx$ is equal to

$\frac{1}{{\sqrt {1 - {x^2}} }}\,\, + \,\,c$

$\frac{{ - \,\,x}}{{\sqrt {1\,\, - \,\,{x^2}} }}\,\, + \,\,c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

\int \tan \,({\sin ^{ - 1}}\,x)\,\;dx

is equal to

View answer and explanation

Correct option: D

- \,\sqrt {1\,\, - \,\,{x^2}} + \,\,c

Option D (

\ - \,\sqrt {1\\,\\ - \\,{x^2}} + \\,\c

) is correct because the integral of

\tan(\sin^{-1}x)

can be solved by trigonometric substitution and simplification.

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