JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 17)
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 :
where C is a constant of integration, then the ordered pair (A(x), B(x)) can be :
(x + 1, -$${\sqrt x }$$)
(x + 1, $${\sqrt x }$$)
(x - 1, -$${\sqrt x }$$)
(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) = –$${\sqrt x }$$
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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