The value of $\left[ {\mathop \smallint \nolimits_0^{{{\sin }^2}\theta } {{\sin }^{ - 1}}\,\,\sqrt {\phi \,} d\phi \,\, + \,\,\mathop \smallint \nolimits_0^{{{\cos }^2}\theta } \,\,{{\cos }^{ - 1}}\,\,\sqrt {\phi \,} d\phi } \right]$ is equal to

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

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Analyzing +4 | -1

The value of

\left[ {\int \nolimits_0^{{{\sin }^2}\theta } {{\sin }^{ - 1}}\,\,\sqrt {\phi \,} d\phi \,\, + \,\,\int \nolimits_0^{{{\cos }^2}\theta } \,\,{{\cos }^{ - 1}}\,\,\sqrt {\phi \,} d\phi } \right]

is equal to

View answer and explanation

Correct option: D

\pi /4

Option D (

\pi/4

) is correct because the derivative of the function inside the brackets is zero, making the function constant. Evaluating at

\theta = \pi/4

gives

\pi/4

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