$\smallint \,\,\cos \,\,\left\{ {2\,\,{{\tan }^{ - 1}}\,\,\sqrt {\frac{{1 - x}}{{1 + x}}\,} } \right\}\,\,dx$ is equal to

$\frac{1}{8}\,\,\left( {{x^2} - 1} \right) + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \,\,\cos \,\,\left\{ {2\,\,{{\tan }^{ - 1}}\,\,\sqrt {\frac{{1 - x}}{{1 + x}}\,} } \right\}\,\,dx

is equal to

View answer and explanation

Correct option: D

\frac{{{x^2}}}{2} + c

Option D (

\frac{{{x^2}}}{2} + c

) is correct because after simplifying the trigonometric expression, the integral becomes

\int x dx

, which evaluates to

\frac{x^2}{2} + c