Question

The value of the integral ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{3\sqrt{\cos \theta }}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^5}\mathrm{d}\theta$ equals

Answer 1

Evaluate the given integral.

Given:${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{3\sqrt{\cos \theta }}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^5}\mathrm{d}\theta$

Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{3\sqrt{\cos \theta }}{{\left(\sqrt{\sin \theta }+\sqrt{\cos \theta }\right)}^5}d\theta$

$$\begin{gathered} \Rightarrow I=3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cos \theta }}{{\left(\sqrt{\sin \theta }+\sqrt{\cos \theta }\right)}^5}\mathrm{d}\theta ...\left(i\right) \\ \Rightarrow I=3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin \theta }}{{\left(\sqrt{\cos \theta }+\sqrt{\sin \theta }\right)}^5}\mathrm{d}\theta ...\left(ii\right)\left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(x-a)dx\right] \end{gathered}$$

By adding $\left(i\right)$ and $\left(ii\right)$, we get

$$\begin{gathered} \therefore 2I=3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cos \theta }+\sqrt{\sin \theta }}{{\left(\sqrt{\cos \theta }+\sqrt{\sin \theta }\right)}^5}d\theta \\ \Rightarrow 2I=3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{{\left(\sqrt{\cos \theta }+\sqrt{\sin \theta }\right)}^4}\mathrm{d}\theta \\ \Rightarrow 2I=3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{{\left(\sqrt{\cos \theta }\right)}^4{\left(1+{\displaystyle \frac{\sqrt{\sin \theta }}{\sqrt{\cos \theta }}}\right)}^4}\mathrm{d}\theta \\ \Rightarrow 2I=3{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{{\left(\cos \left(\theta \right)\right)}^2{\left(1+{\displaystyle \frac{\sqrt{\sin \theta }}{\sqrt{\cos \theta }}}\right)}^4}\mathrm{d}\theta \\ \Rightarrow \frac{2I}{3}={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{se{c}^2\theta }{{\left(\sqrt{\tan \theta }+1\right)}^4}\mathrm{d}\theta \left[\because \frac{\sin \left(x\right)}{\cos \left(x\right)}=\tan \left(x\right),\cos \left(x\right)=\frac{1}{\sec \left(x\right)}\right] \end{gathered}$$

Let $\tan \theta =t^2$

$\therefore se{c}^2\theta \mathrm{d}\theta =2t\mathrm{d}t$

$$\begin{gathered} \Rightarrow \frac{2I}{3}={\int }_0^{\infty }\frac{2t}{{(t+1)}^4}\mathrm{d}t \\ \Rightarrow \frac{I}{3}={\int }_0^{\infty }\frac{t}{{(t+1)}^4}\mathrm{d}t \\ \Rightarrow \frac{I}{3}={\int }_0^{\infty }\left[\frac{1}{{(t+1)}^3}-\frac{1}{{(t+1)}^4}\right]\mathrm{d}t \\ \Rightarrow I={\left|\frac{-3}{2{(t+1)}^2}+\frac{1}{{(t+1)}^3}\right|}_0^{\infty } \\ \Rightarrow I=\frac{3}{2}-1 \\ \Rightarrow I=\frac{1}{2} \end{gathered}$$

Hence, the value of the integral ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{3\sqrt{\cos \theta }}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^5}d\theta$ equals $\frac{1}{2}$.