Question Number 192349 by 073 last updated on 15/May/23
Answered by mehdee42 last updated on 15/May/23
$${I}=\int\frac{{cos}^{\mathrm{4}} {x}\sqrt{\mathrm{1}+{sinx}}}{{cosx}}\:{dx}=\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\sqrt{\mathrm{1}+{sinx}}\:{cosxdx} \\ $$$${let}\::\:\mathrm{1}+{sinx}={u}^{\mathrm{2}} \:\Rightarrow{cosxdx}=\mathrm{2}{udu} \\ $$$$\Rightarrow{I}=\mathrm{2}\int\left(−{u}^{\mathrm{2}} +\mathrm{2}{u}\right){u}^{\mathrm{2}} {du}\:=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{4}} \right)+{c} \\ $$$$\Rightarrow{I}=\left(\mathrm{1}+{sinx}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}\sqrt{\mathrm{1}+{sinx}}}{\mathrm{5}}\right)+{c} \\ $$