Question

$I=\int \frac{(1+{\tan }^{2}x{)}^2}{1+{\tan }^{2}x}{\sec }^{2}xdx=$

Answer 1

$I=\int \frac{(1+ta{n}^{2}x{)}^2}{1+ta{n}^{2}x}se{c}^{2}xdx$

let $t=tanx$

$dt=se{c}^{2}xdx$

$\Rightarrow \int \frac{(1+t^2{)}^2}{(1+t^2)}dt=\int (1+t^2)dt=\frac{t+t^3+c}{3}$

$\Rightarrow tanx+\frac{(tanx{)}^3}{3}+c$ Solution