QJEEM20Ma2ae2: JEE Main 2020 Mathematics - Class 12 JEEM20Ma2ae2 Solution: If = A(x) + B(x) + C, where C is a constant of integration,…

Question

Verified Solution • 2025–26 Session moderate

If

\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx

= A(x)

{\tan ^{ - 1}}\left( {\sqrt x } \right)

+ B(x) + C,
where C is a constant of integration, then the ordered pair (A(x), B(x)) can be :

Accepted Answer

Given, I =

\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx

Let

{\sin ^{ - 1}}\left( {{{\sqrt x } \over {\sqrt {1 + x} }}} \right)

=

\theta

\Rightarrow

{{{\sqrt x } \over {\sqrt {1 + x} }} = \sin \theta }

\Rightarrow

tan

\theta

=

{{{\sqrt x } \over 1}}

\Rightarrow

\theta

=

{\tan ^{ - 1}}\left( {\sqrt x } \right)

\therefore

I =

\int {{{\tan }^{ - 1}}\left( {\sqrt x } \right)dx}

=

\int {{{\tan }^{ - 1}}\left( {\sqrt x } \right).1dx}

Applying integration by parts,

I =

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{1 \over {1 + x}}{1 \over {2\sqrt x }}xdx}

Let

{\sqrt x }

= t

\Rightarrow

x = t²

\Rightarrow

dx = 2tdt

\therefore

I =

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{{{t^2}} \over {\left( {1 + {t^2}} \right)\left( {2t} \right)}}2tdt}

=

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{{\left( {{t^2} + 1} \right) - 1} \over {\left( {1 + {t^2}} \right)}}dt}

=

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - t + {\tan ^{-1}}t + c

=

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \sqrt x + {\tan ^{ - 1}}\sqrt x + c

\therefore

A(x) = x + 1, B(x) = –

{\sqrt x }

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