$\smallint \,\,\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\;\,\,dx$ is equal to

$\log \,\,({\sin ^{ - 1}}x) + c$

$\log \,\,\left( {\sqrt {1 - {x^2}} } \right) + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \,\,\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\;\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}\,\,{\left( {{{\sin }^{ - 1}}x} \right)^2} + c

Option B (

\frac{1}{2}\,{\left( {{{\sin }^{ - 1}}x} \right)^2} + c

) is correct because it results from a straightforward u-substitution with

u = \sin^{-1}x