Question Number 5262 by Kasih last updated on 03/May/16
$$\int\mathrm{cos}^{\mathrm{3}} {x}\sqrt{\mathrm{sin}{x}}{dx} \\ $$
Answered by Yozzii last updated on 03/May/16
$${J}=\int{cos}^{\mathrm{3}} {x}\sqrt{{sinx}}{dx}=\int{cosxcos}^{\mathrm{2}} {x}\sqrt{{sinx}}{dx} \\ $$$${J}=\int{cosx}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\sqrt{{sinx}}{dx}. \\ $$$${Let}\:{u}={sinx}\Rightarrow{du}={cosxdx}. \\ $$$$\therefore{J}=\int\left(\mathrm{1}−{u}^{\mathrm{2}} \right){u}^{\mathrm{1}/\mathrm{2}} {du} \\ $$$${J}=\int\left({u}^{\mathrm{1}/\mathrm{2}} −{u}^{\mathrm{5}/\mathrm{2}} \right){du} \\ $$$${J}=\frac{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{u}^{\mathrm{7}/\mathrm{2}} }{\mathrm{7}}+{K} \\ $$$${J}=\frac{\mathrm{2}{sin}^{\mathrm{3}/\mathrm{2}} {x}}{\mathrm{3}}−\frac{\mathrm{2}{sin}^{\mathrm{7}/\mathrm{2}} {x}}{\mathrm{7}}+{K} \\ $$