JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 7)
The value of the integral
$$\int_0^{1/2} {{{1 + \sqrt 3 } \over {{{({{(x + 1)}^2}{{(1 - x)}^6})}^{1/4}}}}dx} $$ is ........
$$\int_0^{1/2} {{{1 + \sqrt 3 } \over {{{({{(x + 1)}^2}{{(1 - x)}^6})}^{1/4}}}}dx} $$ is ........
Answer
2
Explanation
Let $$I = \int_0^{1/2} {{{1 + \sqrt 3 } \over {{{[{{(x + 1)}^2}{{(1 - x)}^6}]}^{1/4}}}}dx} $$
$$ \Rightarrow I = \int_0^{1/2} {{{1 + \sqrt 3 } \over {{{(1 - x)}^2}{{\left[ {{{\left( {{{1 - x} \over {1 + x}}} \right)}^6}} \right]}^{1/4}}}}dx} $$
Put $${{1 - x} \over {1 + x}} = t \Rightarrow {{ - 2dx} \over {{{(1 + x)}^2}}} = dt$$
when x = 0, t = 1, x = $${1 \over 2}$$, t = $${1 \over 3}$$
$$ \therefore $$ $$I = \int_1^{1/3} {{{(1 + \sqrt 3 )dt} \over { - 2{{(t)}^{6/4}}}}} $$
$$ \Rightarrow I = {{ - (1 + \sqrt 3 )} \over 2}\left[ {{{ - 2} \over {\sqrt t }}} \right]_1^{1/3}$$
$$ \Rightarrow I = (1 + \sqrt 3 )(\sqrt 3 - 1) \Rightarrow I = 3 - 1 = 2$$
$$ \Rightarrow I = \int_0^{1/2} {{{1 + \sqrt 3 } \over {{{(1 - x)}^2}{{\left[ {{{\left( {{{1 - x} \over {1 + x}}} \right)}^6}} \right]}^{1/4}}}}dx} $$
Put $${{1 - x} \over {1 + x}} = t \Rightarrow {{ - 2dx} \over {{{(1 + x)}^2}}} = dt$$
when x = 0, t = 1, x = $${1 \over 2}$$, t = $${1 \over 3}$$
$$ \therefore $$ $$I = \int_1^{1/3} {{{(1 + \sqrt 3 )dt} \over { - 2{{(t)}^{6/4}}}}} $$
$$ \Rightarrow I = {{ - (1 + \sqrt 3 )} \over 2}\left[ {{{ - 2} \over {\sqrt t }}} \right]_1^{1/3}$$
$$ \Rightarrow I = (1 + \sqrt 3 )(\sqrt 3 - 1) \Rightarrow I = 3 - 1 = 2$$
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