$\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}$ is equal to

Let, $\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}$

On taking sin x as first function and e^2x as second function and integrating by parts, we get

$\begin{array}{l}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}=\sin \mathrm{x}\cdot \int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}-\int \left[\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})\cdot \int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\right]\mathrm{dx}\\ =\sin \mathrm{x}\cdot \frac{{\mathrm{e}}^{2\mathrm{x}}}{2}-\int \cos \mathrm{x}\cdot \frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{dx}\\ =\sin \mathrm{x}\cdot \frac{{\mathrm{e}}^{2\mathrm{x}}}{2}-\frac{1}{2}\int {\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{xdx}\\ \Rightarrow \int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}=\sin \mathrm{x}\cdot \frac{{\mathrm{e}}^{2\mathrm{x}}}{2}-\frac{1}{2}{\mathrm{l}}_1------\left(\mathrm{i}\right)\end{array}$

where ${\mathrm{I}}_1=\int {\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{xdx}$

On taking cos x as first function and e^2x as second function and integrating by parts, we get

$\begin{array}{l}{\mathrm{I}}_1=\cos \mathrm{x}\cdot \int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}-\int \left[\frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})\cdot \int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\right]\mathrm{dx}\\ =\cos \mathrm{x}\cdot \frac{{\mathrm{e}}^{2\mathrm{x}}}{2}-\int (-\sin \mathrm{x})\cdot \frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\mathrm{dx}+{\mathrm{C}}_1\\ =\frac{{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}}{2}+\frac{1}{2}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}+{\mathrm{C}}_1\end{array}$

On putting value of I₁ in Eq. (i), we get

$\begin{array}{l}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}=\frac{{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}}{2}-\frac{1}{2}\left[\frac{{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}}{2}+\frac{1}{2}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}+{\mathrm{C}}_1\right]\\ \Rightarrow \int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}=\frac{{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}}{2}-\frac{{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}}{4}-\frac{1}{4}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}-\frac{1}{2}{\mathrm{C}}_1\\ \Rightarrow \int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}+\frac{1}{4}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}\\ =\frac{{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}}{2}-\frac{{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}}{4}-\frac{1}{2}{\mathrm{C}}_1\\ \Rightarrow \frac{5}{4}\int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}=\frac{{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}}{2}-\frac{{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}}{4}-\frac{1}{2}{\mathrm{C}}_1\\ \Rightarrow \int {\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{xdx}=\frac{4}{5}\left[\frac{{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}}{2}-\frac{{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}}{4}-\frac{1}{2}{\mathrm{C}}_1\right]\\ =\frac{2}{5}{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}-\frac{1}{5}{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}-\frac{2}{5}{\mathrm{C}}_1\\ =\frac{2}{5}{\mathrm{e}}^{2\mathrm{x}}\sin \mathrm{x}-\frac{1}{5}{\mathrm{e}}^{2\mathrm{x}}\cos \mathrm{x}+\mathrm{C}\end{array}$

$\begin{array}{l}\mathrm{where} ,\mathrm{C}=-\frac{2}{5}{\mathrm{C}}_1\\ \mathrm{Hence} , \mathrm{I}=\frac{{\mathrm{e}}^{2\mathrm{x}}}{5}(2\sin \mathrm{x}-\cos \mathrm{x}+\mathrm{C})\end{array}$