Question 223 of 416MediumMCQ
If ∫1+xsin−1xdx=A(x)tan−1(x)+B(x)+C, where C is a constant of integration, then the ordered pair (A(x),B(x)) can be :
Solutions · 1
Official solution
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) = --\[{\sqrt x }\]
Answer the question to unlock