Mathematics - Definite Integration Question with Solution | TestHub

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3\sqrt{\cos \theta }}{{\left(\sqrt{\sin \theta }+\sqrt{\cos \theta }\right)}^5}.d\theta$…(i)
$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3\sqrt{\sin \theta }}{{\left(\sqrt{\cos \theta }+\sqrt{\sin \theta }\right)}^5}.d\theta$…(ii)$\mathrm{using}\mathrm{ }\underset{a}{\overset{b}{\int }}f\left(x\right).dx=\underset{a}{\overset{b}{\int }}f\left(a+b-x\right).dx$
(i)$+$(ii)
$\Rightarrow 2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3\left(\sqrt{\sin \theta }+\sqrt{\cos \theta }\right)}{{\left(\sqrt{\sin \theta }+\sqrt{\cos \theta }\right)}^5}.d\theta$
$\Rightarrow 2I=3\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{d\theta }{{\left(\sqrt{\sin \theta }+\sqrt{\cos \theta }\right)}^4}$
$\Rightarrow \frac{2I}{3}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sec }^2\theta .d\theta }{{\left(1+\sqrt{\tan \theta }\right)}^4}$
Put$\tan \theta =t^2$
${\sec }^2\theta .d\theta =2t.dt$
when
$\begin{array}{cc}\theta \to 0,& t\to 0\end{array}$
$\begin{array}{cc}\theta \to \frac{\pi }{2},& t\to \infty \end{array}$
$\Rightarrow \frac{2I}{3}=\underset{0}{\overset{\infty }{\int }}\frac{2tdt}{{\left(1+t\right)}^4}$
$\Rightarrow \frac{I}{3}=\underset{0}{\overset{\infty }{\int }}\frac{\left(t+1\right)-1}{{\left(1+t\right)}^4}.dt$
$\frac{I}{3}=\underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(1+t\right)}^3}-\frac{1}{{\left(1+t\right)}^4}.dt$
$\frac{I}{3}={\left[-\frac{1}{2{\left(1+t\right)}^2}+\frac{1}{3{\left(1+t\right)}^3}\right]}_0^{\infty }$
$\frac{I}{3}=\left[0+0-\left(-\frac{1}{2}+\frac{1}{3}\right)\right]$
$\frac{I}{3}=\frac{1}{6}\Rightarrow I=\frac{1}{2}$