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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 :

JEE · Math · previous-year question

  1. A.(x + 1, -$${\sqrt x }$$)correct
  2. B.(x + 1, $${\sqrt x }$$)
  3. C.(x - 1, -$${\sqrt x }$$)
  4. D.(x - 1, $${\sqrt x }$$)

Answer

A. (x + 1, -$${\sqrt x }$$)

Explanation

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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