The value of the integral $\int\limits_{ - 1}^1 {{\rm{ }}{{\log }_e}} \left( {\sqrt {1 - {\rm{ }}x} + \sqrt {1 + {\rm{ }}x} } \right)dx$ is equal to

${\rm{lo}}{{\rm{g}}_e}2 + \frac{\pi }{2} - 1$

$\frac{1}{2}{\rm{lo}}{{\rm{g}}_e}2 + \frac{\pi }{4} - \frac{3}{2}$

${\rm{2lo}}{{\rm{g}}_e}2 + \frac{\pi }{4} - 1$

${\rm{2lo}}{{\rm{g}}_e}2 + \frac{\pi }{2} - \frac{1}{2}$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The value of the integral

\int\limits_{ - 1}^1 {\mathrm{ }{{\log }_e}} \left( {\sqrt {1 - \mathrm{ }x} + \sqrt {1 + \mathrm{ }x} } \right)dx

is equal to

View answer and explanation

Correct option: C

\mathrm{lo}{\mathrm{g}_e}2 + \frac{\pi }{2} - 1

Option C (

\mathrm{lo}{\mathrm{g}_e}2 + \frac{\pi }{2} - 1

) is correct because it is the result of evaluating the definite integral using integration by parts and trigonometric substitution.