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QJEEAdv18Ma9658: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma9658 Solution: The value of the integral is ........

The value of the integral

∫01/21+3((x+1)2(1−x)6)1/4dx\int_0^{1/2} {{{1 + \sqrt 3 } \over {{{({{(x + 1)}^2}{{(1 - x)}^6})}^{1/4}}}}dx} is ........

Step-by-Step Verified Solution

Let I=∫01/21+3[(x+1)2(1−x)6]1/4dxI = \int_0^{1/2} {{{1 + \sqrt 3 } \over {{{[{{(x + 1)}^2}{{(1 - x)}^6}]}^{1/4}}}}dx}

⇒I=∫01/21+3(1−x)2[(1−x1+x)6]1/4dx\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−x1+x=t⇒−2dx(1+x)2=dt{{1 - x} \over {1 + x}} = t \Rightarrow {{ - 2dx} \over {{{(1 + x)}^2}}} = dt

when x = 0, t = 1, x = 12{1 \over 2} , t = 13{1 \over 3}

∴\therefore I=∫11/3(1+3)dt−2(t)6/4I = \int_1^{1/3} {{{(1 + \sqrt 3 )dt} \over { - 2{{(t)}^{6/4}}}}}

⇒I=−(1+3)2[−2t]11/3\Rightarrow I = {{ - (1 + \sqrt 3 )} \over 2}\left[ {{{ - 2} \over {\sqrt t }}} \right]_1^{1/3}

⇒I=(1+3)(3−1)⇒I=3−1=2\Rightarrow I = (1 + \sqrt 3 )(\sqrt 3 - 1) \Rightarrow I = 3 - 1 = 2

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