Question

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

Answer 1

$I=\int\limits_{0}^{\infty} \frac{t^{4} d t}{\left(1+t^{2}\right)^{6}}$
Let $t=\tan \theta$
$d t=\sec ^{2} \theta\, d \theta$
$=\int\limits_{0}^{\pi / 2} \frac{\tan ^{4} \theta \sec ^{2} \theta \,d \theta}{\sec ^{12} \theta}=\int\limits_{0}^{\pi / 2} \sin ^{4} \theta \cdot \cos ^{6} \theta\, d \theta$
$=\frac{(3.1) \cdot(5.3 .1)}{(10.8 .6 .4 .2)} \cdot \frac{\pi}{2}=\frac{3 \pi}{512}=\frac{3 \pi}{64 k}$
$\Rightarrow k=8$