QJEEAdv19Ma4a0f: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma4a0f Solution: The value of the integral equals ..............

The given integral

$I = \int\limits_0^{\pi /2} {{{3\sqrt {\cos \theta } } \over {{{(\sqrt {\cos \theta } + \sqrt {\sin \theta } )}^5}}}} d\theta$ ....(i)

$\Rightarrow I = \int\limits_0^{\pi /2} {{{3\sqrt {\sin \theta } } \over {{{(\sqrt {\sin \theta } + \sqrt {\cos \theta } )}^5}}}} d\theta$ ....(ii)

[Using the property $\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} }$ ]

Now, on adding integrals (i) and (ii), we get

$2I = \int\limits_0^{\pi /2} {{3 \over {{{(\sqrt {\sin \theta } + \sqrt {\cos \theta } )}^4}}}} d\theta$

$= \int\limits_0^{\pi /2} {{{3{{\sec }^2}\theta } \over {{{(1 + \sqrt {\tan \theta } )}^4}}}} d\theta$

Now, let $\tan \theta = {t^2} \Rightarrow {\sec ^2}\theta \,d\theta = 2t\,dt$

and at $\theta = {\pi \over 2}$ , t $\to$ $\infty$

and at $\theta = 0$ , t $\to$ 0

So, $2I = \int_0^\infty {{{6\,t\,dt} \over {{{(1 + t)}^4}}} = 6} \int_0^\infty {{{t + 1 - 1} \over {{{(t + 1)}^4}}}} dt$

$\Rightarrow I = 3\left[ {\int_0^\infty {{{dt} \over {{{(t + 1)}^3}}} - } \int_0^\infty {{{dt} \over {{{(t + 1)}^4}}}} } \right]$

$= 3\left[ { - {1 \over {2{{(t + 1)}^2}}} + {1 \over {3{{(t + 1)}^3}}}} \right]_0^\infty$

$\Rightarrow I = 3\left[ {{1 \over 2} - {1 \over 3}} \right] = 3\left( {{1 \over 6}} \right) = {1 \over 2} \Rightarrow I = 0.5$