Question

The value of the integral $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3\sqrt{\cos \theta }}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^5}d\theta$ equals to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{2}$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3\left(\sqrt{\cos \theta }\right)}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^5}d\theta \cdots \left(i\right)$
$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3\left(\sqrt{\sin \theta }\right)}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^5}d\theta \cdots \left(ii\right)$
Adding equation $\left(i\right)$ and $\left(ii\right),$ we get
$2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3}{{(\sqrt{\cos \theta }+\sqrt{\sin \theta })}^4}d\theta$
$2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{3{\sec }^2\theta }{{(1+\sqrt{\tan \theta })}^4}d\theta$
Let $1+\sqrt{\tan \theta }=t\Rightarrow \tan \theta =(t-1{)}^2{\sec }^2\theta d\theta =2(t-1)dt$

$2I=\underset{1}{\overset{\infty }{\int }}\frac{3\times 2(t-1)}{(t{)}^4}dt$
$I=3\underset{1}{\overset{\infty }{\int }}(t^{-3}-t^{-4})dt$
$I=3{\left[\frac{t^{-2}}{-2}-\frac{t^{-3}}{-3}\right]}_1^{\infty }=3\left[\right(0-0)-(-\frac{1}{2}+\frac{1}{3})]I=\frac{1}{2}$