Question

If $\underset{0}{\overset{\infty }{\int }}\frac{dx}{x^{3/2}+1}=\frac{a\pi }{b^{3/2}},$ where $\text{gcd}(a,b)=1,$ then

• A. $a=2$
• B. $b=3$
• C. $a=4$
• D. $b=5$

Answer 1

Answer: (B), (C)

$a=4$

$I=\underset{0}{\overset{\infty }{\int }}\frac{dx}{x^{3/2}+1}$
Put $x={\tan }^2\theta$
$\Rightarrow dx=2\tan \theta {\sec }^2\theta d\theta$
$I=\underset{0}{\overset{\pi /2}{\int }}\frac{2\tan \theta {\sec }^2\theta d\theta }{1+{\tan }^3\theta }$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{2\sin \theta d\theta }{{\sin }^3\theta +{\cos }^3\theta }\cdots \left(1\right)$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{2\cos \theta d\theta }{{\sin }^3\theta +{\cos }^3\theta }\cdots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{(\sin \theta +\cos \theta )d\theta }{{\sin }^3\theta +{\cos }^3\theta }$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{(\sin \theta +\cos \theta )d\theta }{(\sin \theta +\cos \theta )({\sin }^2\theta +{\cos }^2\theta -\sin \theta \cos \theta )}$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{d\theta }{1-\sin \theta \cos \theta }$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{{\sec }^2\theta d\theta }{1+{\tan }^2\theta -\tan \theta }$
Again, put $\tan \theta =t\Rightarrow {\sec }^2\theta d\theta =dt$
$I=\underset{0}{\overset{\infty }{\int }}\frac{dt}{t^2-t+1}$
$=\underset{0}{\overset{\infty }{\int }}\frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$={\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)|}_0^{\infty }$
$=\frac{2}{\sqrt{3}}(\frac{\pi }{2}-\frac{-\pi }{6})$
$=\frac{4\pi }{3^{3/2}}$