Question

$\int \sqrt{\sin x}{\cos }^{3}xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \sqrt{\sin x}·{\cos }^{3}xdx \\ =\int \sqrt{\sin x}·\left({\cos }^{2}x\right)·\cos xdx \\ =\int \sqrt{\sin x}\left(1-{\sin }^{2}x\right)·\cos xdx \\ \mathrm{Putting}\sin x=t \\ \Rightarrow \cos xdx=dt \\ \therefore I=\int \sqrt{t}\left(1-t^2\right)·dt \\ =\int t^{\frac{1}{2}}dt-\int t^{\frac{1}{2}}·t^{2}dt \\ =\int t^{\frac{1}{2}}dt-\int t^{\frac{5}{2}}dt \\ =\frac{t^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{t^{{\displaystyle \frac{7}{2}}}}{{\displaystyle \frac{7}{2}}}+C \\ =\frac{2}{3}{t}^{\frac{3}{2}}-\frac{2}{7}{t}^{\frac{7}{2}}+C \\ =\frac{2}{3}{\sin }^{\frac{3}{2}}x-\frac{2}{7}{\sin }^{\frac{7}{2}}x+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}t=\mathit{}\sin \mathit{}x\right] \end{gathered}$$