Answer
\[\frac{e^x}{10}\left[5-\cos 2x-2\sin 2x\right]+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int e^x\sin^2 x\: dx\]
\[I=\int e^x\left[\frac{1-\cos 2x}{2}\right]\: dx\]
\[I=\frac{1}{2}\int e^x \:dx-\frac{1}{2}\int e^x\cos 2x \;dx\;\;\;\;...(1)\]
Let \[I_1=\int e^x\cos 2x \;dx\]
Using integration by parts
\[I_1=\cos 2x\int e^x \;dx-\int \left((\cos 2x)'\int e^x \:dx\right)dx\]
\[I_1=\cos 2x\;(e^x)+2\int(\sin 2x)\;e^{x}\:dx\]
\[I_1=\cos 2x\;(e^x)+2\left[\sin 2x\;e^x-2\int\cos 2x\;e^x\:dx\right]\]
\[I_1=e^x \cos 2x+2e^x\sin 2x-4I_1\]
\[5I_1=e^x(\cos 2x+2\sin 2x)\]
\[I_1=\frac{e^x}{5}(\cos 2x+2\sin 2x)\;\;\;...(2)\]
Using (2) in (1)
\[\Rightarrow I=\frac{e^x}{2}-\frac{e^x}{10}(\cos 2x+2\sin 2x)+C\]
Where $C$ is constant of integration
\[I=\frac{e^x}{10}\left[5-\cos 2x-2\sin 2x\right]+C\]
Hence , \[I=\frac{e^x}{10}\left[5-\cos 2x-2\sin 2x\right]+C\;\]
