Question

\[\int e^{2x} \cos^2 x\ dx\]

Answer 1

\[\text{ Let I }= \int e^{2x} \cos^2 x \text{ dx }\]
\[ = \int e^{2x} \left( \frac{1 + \cos 2x}{2} \right)\text{ dx }\]
\[ = \frac{1}{2}\int e^{2x} \text{ dx }+ \frac{1}{2}\int e^{2x} \text{ cos }\left( 2x \right)dx \]
\[ = \frac{e^{2x}}{4} + \frac{1}{2} I_1 . . . . . \left( 1 \right)\]
\[\text{ Where}\ I_1 = \int e^{2x} \text{ cos  2x  dx}\]
`\text{Considering cos  ( 2x ) as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[ I_1 = \cos \left( 2x \right)\frac{e^{2x}}{2} - \int\left( - 2 \text{ sin  2x } \times \frac{e^{2x}}{2} \right)dx\]
\[ \Rightarrow I_1 = \frac{\text{ cos } \left( 2x \right) e^{2x}}{2} + \int e^{2x} \text{ sin } \left( 2x \right) dx\]
`\text{Considering sin   ( 2x ) as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[ I_1 = \frac{\text{ cos }\left( 2x \right) e^{2x}}{2} + \text{ sin }\left( 2x \right)\frac{e^{2x}}{2} - \int 2 \cos 2x\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \left( \cos 2x + \sin 2x \right)}{2} - I_1 \]
\[ \Rightarrow \text{ 2 }I_1 = \frac{e^{2x} \left( \cos 2x + \sin 2x \right)}{2}\]
\[ \Rightarrow I_1 = \frac{e^{2x} \left( \cos 2x + \sin 2x \right)}{4} . . . . . \left( 2 \right)\]
\[\text{ From }\left( 1 \right) \text{ and }\ \left( 2 \right)\]
\[I = \frac{e^{2x}}{4} + \frac{e^{2x}}{8}\left( \cos 2x + \sin 2x \right) + C\]