Question

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

Answer 1

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

Integrate by parts -

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

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

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

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

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

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

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

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

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

Hence, proved.