JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 19)
The integral $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$ is equal to :
(where C is a constant of integration.)
(where C is a constant of integration.)
$$ - 2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C$$
$$ - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$$
$$ - \sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$$
$$2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C$$
Explanation
I = $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$
= $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} $$
Let 1 + $$\sqrt x $$ = t
$$ \Rightarrow $$$$\,\,\,$$$${1 \over {2\sqrt x }}\,dx$$ = dt
I = $$\int {{{2dt} \over {t\sqrt {2t - {t^2}} }}} $$
Again let t = $${1 \over z}$$
$$ \Rightarrow $$$$\,\,\,$$ dt = $$-$$ $${1 \over {{z^2}}}dz$$
$$\therefore\,\,\,$$ I = 2 $$\int {{{ - {1 \over {{z^2}}}dz} \over {{1 \over z}\sqrt {{2 \over z} - {1 \over {{z^2}}}} }}} $$
= 2$$\int {{{ - dz} \over {\sqrt {2z - 1} }}} $$
= $$ - 2\sqrt {2z - 1} + C$$
= $$ - 2\sqrt {{2 \over t} - 1} + C$$
= $$- 2\sqrt {{{2 - t} \over t}} + C$$
= $$ - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$$
= $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} $$
Let 1 + $$\sqrt x $$ = t
$$ \Rightarrow $$$$\,\,\,$$$${1 \over {2\sqrt x }}\,dx$$ = dt
I = $$\int {{{2dt} \over {t\sqrt {2t - {t^2}} }}} $$
Again let t = $${1 \over z}$$
$$ \Rightarrow $$$$\,\,\,$$ dt = $$-$$ $${1 \over {{z^2}}}dz$$
$$\therefore\,\,\,$$ I = 2 $$\int {{{ - {1 \over {{z^2}}}dz} \over {{1 \over z}\sqrt {{2 \over z} - {1 \over {{z^2}}}} }}} $$
= 2$$\int {{{ - dz} \over {\sqrt {2z - 1} }}} $$
= $$ - 2\sqrt {2z - 1} + C$$
= $$ - 2\sqrt {{2 \over t} - 1} + C$$
= $$- 2\sqrt {{{2 - t} \over t}} + C$$
= $$ - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$$
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