Answer
\[\frac{2}{5}\]
Work Step by Step
Let \[I=\int_{0}^{\frac{π}{2}}\cos ^3 x\sin 2x\; dx\]
\[I=\int_{0}^{\frac{π}{2}}\cos ^3 x(2\sin x\cos x)\; dx\]
\[I=2\int_{0}^{\frac{π}{2}}\cos ^4 x(\sin x)\; dx\;\;\;\ldots (1)\]
Let \[I_1=\int\cos^4 x(\sin x)dx\]
Substitute $\;\cos x=t\;\;\;\ldots (2)$
\[\Rightarrow -\sin xdx=dt\]
\[I_1=-\int t^4 dt=-\frac{t^5}{5}+C\]
Where $C$ is constant of integration
From (2)
\[I_1=-\frac{\cos^5 x}{5}+C\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=2\left[-\frac{\cos^5 x}{5}\right]_{0}^{\frac{π}{2}}=2\left[0+\frac{1}{5}\right]=\frac{2}{5}\]
Hence \[I=\frac{2}{5}\;\;.\]
