If ${I_1} = \smallint {\sin ^{ - 1}}x\,dx$ and ${I_2} = \smallint {\sin ^{ - 1}}\,\,\sqrt {1\,\, - \,\,{x^2}} \;dx$, then

${I_2} = \frac{\pi }{2}\,\,{I_1}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Analyzing +4 | -1

{I_1} = \int {\sin ^{ - 1}}x\,dx

and

{I_2} = \int {\sin ^{ - 1}}\,\,\sqrt {1\,\, - \,\,{x^2}} \;dx

, then

View answer and explanation

Correct option: C

{I_1} + \,\,{I_2} = \frac{\pi }{2}\,\,x

Option C (

{I_1} + \, {I_2} = \frac{\pi }{2}\,x

) is correct because the sum of the integrals simplifies to an integral involving the constant

\frac{\pi}{2}