Question

Let $\underset{0}{\overset{\infty }{\int }}\frac{t^{4}dt}{{\left(1+t^2\right)}^6}=\frac{3\pi }{64k}$ then $k$ is equal to

Answer 1

Answer: 8

$I=\underset{0}{\overset{\infty }{\int }}\frac{t^{4}dt}{{\left(1+t^2\right)}^6}$
Let $t=\tan \theta$
$dt={\sec }^2\theta d\theta$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{{\tan }^4\theta {\sec }^2\theta d\theta }{{\sec }^{12}\theta }=\underset{0}{\overset{\pi /2}{\int }}{\sin }^4\theta \cdot {\cos }^6\theta d\theta$
$=\frac{(3.1)\cdot (\mathrm{5.3.1})}{(\mathrm{10.8.6.4.2})}\cdot \frac{\pi }{2}=\frac{3\pi }{512}=\frac{3\pi }{64k}$
$\Rightarrow k=8$