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

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

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

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