Question

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

Answer 1

\[\text{ Let I }= \int e^{2x} \sin x \text{ cos x dx }\]
\[I = \frac{1}{2}\int e^{2x} \left( 2 \sin x \text{ cos  x }\right)\text{ dx }\]
\[ \Rightarrow I = \frac{1}{2}\int e^{2x} \text{ sin 2x dx }\]
`\text{Considering sin  2x  as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[I = \frac{1}{2}\left[ \sin2x\frac{e^{2x}}{2} - \int2\cos2x\frac{e^{2x}}{2}dx \right]\]
\[ \Rightarrow I = \frac{e^{2x} \sin2x}{4} - \frac{1}{2}\int e^{2x} \cos2xdx\]
\[ \Rightarrow I = \frac{e^{2x} \sin2x}{4} - \frac{1}{2} I_1 . . . . . \left( 1 \right)\]
\[\text{ Where I}_1 = \int e^{2x} \cos2xdx\]
`\text{Considering cos  2x  as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[ I_1 = \cos2x\frac{e^{2x}}{2} - \int - 2 \sin2x\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \cos2x}{2} + \int e^{2x} \sin2x dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \cos2x}{4} + 2I . . . . . \left( 2 \right)\]
\[ \Rightarrow I = \frac{e^{2x} \sin2x}{4} - \frac{1}{2}\left[ \frac{e^{2x} \cos2x}{2} + 2I \right]\]
\[ \Rightarrow I = \frac{e^{2x} \sin2x}{4} - \frac{e^{2x} \cos2x}{4} - \frac{I}{2} \times 2\]
\[ \Rightarrow 2I = \frac{e^{2x} \left( \sin2x - \cos2x \right)}{4} + C\]
\[ \Rightarrow I = \frac{e^{2x}}{8}\left( \sin2x - \cos2x \right) + C\]