Question

$\int {\tan }^{2}x.se{c}^{4}xdx=$

Answer 1

$LetI=\int {\tan }^{2}x.se{c}^{4}xdxPuttanx=t\Rightarrow se{c}^{2}xdx=dt\therefore I=\int t^2(1+t^2)dt=\int (t^2+t^4)dt=\frac{t^3}{3}+\frac{t^5}{5}+C=\frac{ta{n}^{5}x}{5}+\frac{ta{n}^{3}x}{3}+C$