Question Number 60346 by necx1 last updated on 20/May/19
$$\int{x}\mathrm{sec}\:^{\mathrm{3}} {xdx} \\ $$$${please}\:{help} \\ $$
Commented by kaivan.ahmadi last updated on 20/May/19
$${first}\:\int{sec}^{\mathrm{3}} {x}\:{dx}=\left(\frac{{tgx}}{{cosx}}+{ln}\mid{secx}+{tgx}\mid\right)/\mathrm{2} \\ $$$${now}\:{let} \\ $$$${u}={x}\Rightarrow{du}={dx} \\ $$$${dv}={sec}^{\mathrm{3}} {xdx}\Rightarrow{v}=\left(\frac{{tgx}}{{cosx}}+{ln}\mid{secx}+{tgx}\mid\right)/\mathrm{2} \\ $$$${uv}−\int{vdu}={x}\left(\frac{{tgx}}{{cosx}}+{lb}\mid{secx}+{tgx}\mid\right)/\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{{tgx}}{{cosx}}+{ln}\mid{secx}+{tgx}\mid\right){dx} \\ $$$${on}\:{the}\:{other}\:{hand} \\ $$$$\int\frac{{tgx}}{{cosx}}{dx}=\int\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}{dx}=\frac{\mathrm{1}}{{cosx}}={secx} \\ $$$${and} \\ $$$$\int{ln}\mid{secx}+{tgx}\mid{dx}={xln}\left({secx}+{tgx}\right)−{secx}−{tgx} \\ $$