Question

Solve $\int e^{ax}\sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)$

Answer 1

$I=\int e^{ax}\sin bxdx$

Integrate by parts -

$=e^{ax}\int \sin bxdx-\int \frac{d{e}^{ax}}{dx}(\int \sin bxdx)dx$

$=e^{ax}\frac{-cosbx}{b}-\int a{e}^{ax}\frac{-\cos bx}{b}dx$

$=-e^{ax}\frac{\cos bx}{b}+\frac{a}{b}\int e^{ax}\cos bxdx$

$=\frac{e^{ax}}{b}\cos bx+\frac{a}{b}[e^{ax}\int \cos bxdx-\int \frac{d{e}^{ax}}{dx}(\int \cos bxdx)dx]$

$=\frac{e^{ax}}{b}\cos bx+\frac{a}{b}[e^{ax}\frac{\sin bx}{b}-\int a{e}^{ax}\sin bxdx]$

$=\frac{-e^{ax}}{b}\cos bx+\frac{a{e}^{ax}}{b^2}\sin bx-\frac{a^2}{b^2}\int e^{ax}\sin bxdx$

$\Rightarrow I=e^{ax}\frac{-\cos bx}{b}+\frac{a{e}^{ax}}{b^2}\sin bx-I\left(\frac{a^2}{b^2}\right)$

$\Rightarrow I+\frac{a^2}{b^2}I=e^{ax}\frac{(a\sin bx-b\cos bs)}{b^2}$

$\Rightarrow I\frac{a^2+b^2}{b^2}=e^{ax}\frac{(a\sin bx-b\cos bx)}{b^2}$

$\Rightarrow I=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)$

Hence, proved.