Question

$\int {\tan }^{3}2x\sec 2xdx=\frac{1}{k}({\sec }^{3}2x-3\sec 2x).$ Find the value of $k$.

Answer 1

Let $I=\int {\tan }^{3}2x\sec 2xdx=\int {\tan }^{2}2x\tan 2x\sec 2xdx$
$=\int ({\sec }^{2}2x-1)\sec 2x\tan 2xdx$
Put $\sec 2x=t\Rightarrow 2\sec 2x\tan 2xdx=dt$
Therefore
$I=\frac{1}{2}\int (t^2-1)dt=\frac{1}{2}(\frac{t^3}{3}-t)=\frac{1}{6}({\sec }^{3}2x-3\sec 2x)$

So, $k=6$