Question

Integrate the function.
$\int e^{2x}sinxdx.$

Answer 1

Let $I=\int e^{2x}sinxdx$
On taking sin x as first function and $e^{2x}$ as second function and integrating by parts, we get
$\int e^{2x}sinxdx=sinx\int e^{2x}dx-\int \left[\frac{d}{dx}\right(sinx).\int e^{2x}dx]dx=sinx.\frac{e^{2x}}{2}-\int cosx.\frac{e^{2x}}{2}dx=sinx.\frac{e^{2x}}{2}-\frac{1}{2}\int e^{2x}cosxdx\Rightarrow \int e^{2x}sinxdx=sinx.\frac{e^{2x}}{2}-\frac{1}{2}{I}_1........\left(i\right)$
where, $I_1=\int e^{2x}cosxdx$
On taking cos x as first function and $e^{2x}$ as second function and integrating by. parts we get
$I_1=cosx.\int e^{2x}dx-\int \left[\frac{d}{dx}\right(cosx).\int e^{2x}dx]dx=cosx.\frac{e^{2x}}{2}-\int (-sinx).\frac{e^{2x}}{2}dx+C_1=\frac{e^{2x}cosx}{2}+\frac{1}{2}\int e^{2x}sinxdx+C_1$
Putting value of $I_1$ in Eq. (i), we get
$\int e^{2x}sinxdx=\frac{e^{2x}sinx}{2}-\frac{1}{2}[\frac{e^{2x}cosx}{2}+\frac{1}{2}\int e^{2x}sinxdx+C_1]\Rightarrow \int e^{2x}sinxdx=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}\int e^{2x}sinxdx-\frac{1}{2}{C}_1\Rightarrow \int e^{2x}sinxdx+\frac{1}{4}\int e^{2x}sinxdx=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{2}{C}_1\Rightarrow \frac{5}{4}\int e^{2x}sinxdx=\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{2}{C}_1\Rightarrow \int e^{2x}sinxdx=\frac{4}{5}[\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{2}{C}_1]=\frac{2}{5}{e}^{2x}sinx-\frac{1}{5}{e}^{2x}cosx-\frac{2}{5}{C}_1=\frac{2}{5}{e}^{2x}sinx-\frac{1}{5}{e}^{2x}cosx+C$
where, $C=-\frac{2}{5}{C}_1$
$=\frac{e^{2x}}{5}(2sinx-cosx)+C$