Question

$\int {tan}^{2}2xdx$

Answer 1

$\int {\tan }^{2}2xdx$

${\tan }^{2}x={\sec }^{2}x-1$

we have

$\int ({\sec }^{2}2x-1)dx$

$=\int \left({\sec }^{2}2x\right)dx-\int dx$

Let u$=\tan 2x$ i.e., du$=2{\sec }^2\left(2x\right)dx$

$=\frac{1}{2}\int 2\left({\sec }^{2}2x\right)dx-x$

$=\frac{1}{2}\int du-x$

$=\frac{1}{2}u-x+c$

$=\frac{1}{2}\tan 2x-x+c$.