Question

\[\int e^x \sin^2 x\ dx\]

Answer 1

\[\text{ Let I }= \int e^x \text{ sin}^2 x \text{ dx }\]
\[ = \int e^x \left( \frac{1 - \cos 2x}{2} \right)dx\]
\[ = \frac{1}{2}\int e^x dx - \frac{1}{2}\int e^x \text{  cos 2x  dx }\]
\[ = \frac{e^x}{2} - \frac{1}{2}\int e^x \text{ cos }\left( \text{ 2x}\right) dx . . . . . \left( 1 \right)\]
\[\text{ Let I}_1 = \int e^x \text{ cos} \left( 2x \right)dx\]
`\text{Considering cos  ( 2x ) as first function and` `\text{ e}^{t}`   ` \text{ as second function} `
\[ I_1 = \text{ cos } \left( 2x \right) e^x - \int - 2 \text{ sin }\left( 2x \right) e^x dx\]
\[ \Rightarrow I_1 = \text{ cos } \left( 2x \right) e^x + 2\int \text{ sin }\left( 2x \right) e^x dx\]
\[ \Rightarrow I_1 = \text{ cos } \left( 2x \right) e^x + 2\left[ \text{ sin } \left( 2x \right) e^x - \int 2 \text{ cos } \left( 2x \right) e^x dx \right]\]
\[ \Rightarrow I_1 = \text{ cos }\left( 2x \right) e^x + 2 \text{ sin }\left( 2x \right) e^x - 4 I_1 \]
\[ \Rightarrow 5 I_1 = e^x \left( \text{ cos }2x + 2 \text{ sin }2x \right)\]
\[ \Rightarrow I_1 = \frac{e^x}{5}\left( \text{ cos }2x + 2 \text{ sin }2x \right) + C . . . . . \left( 2 \right)\]
`  \text{ From ( 1 ) and ( 2) `
\[I = \frac{e^x}{2} - \frac{e^x}{10}\left( \text{ cos }2x + 2 \text{ sin }2x \right) + C\]