$\int_{0}^{\infty}{e^{-2x}{cos}4xdx}=?$

ধরি, $I=\int{e^{-2x}\cos{4xdx}} =\cos{4x}\int{e^{-2x}dx-\int\left[\frac{d}{dx}\left(\cos{4x}\right)\int{e^{-2x}dx}\right]}dx$

$=\frac{e^{-2x}\cos{4}x}{-2}-\int\left[-4\sin{4x}\times\frac{e^{-2x}}{-2}\right]dx =-\frac{e^{-2x}\cos{4}x}{2}-2\int{e^{-2x}\sin{4xdx}}$

$=\frac{-e^{-2x}\cos{4}x}{2}-2\left[\sin{4x}\int{e^{-2x}dx-\int\left\{\frac{d}{dx}\left(\sin{4x}\right)\int{e^{-2x}dx}\right\}}dx\right]$

$=-\frac{e^{-2x}\cos{4}x}{2}-2\sin{4x}\frac{e^{-2x}}{-2}+2\int{4\cos{4x}\frac{e^{-2x}}{-2}}dx \Rightarrow5I=e^{-2x}\sin{4x}-\frac{e^{-2x}\cos{4x}}{2}$

$\therefore I=\frac{e^{-2x}\sin{4x}}{5}-\frac{e^{-2x}\cos{4x}}{10} \therefore\ \int_{0}^{\infty}e^{-2x}\ cos\,4x\,dx=\left[\frac{e^{-2x}\ sin\ 4x}{5}-\frac{e^{-2x}\ cos\ 4x}{10}\right]_0^\infty$

$=0-0-0+\frac{1}{10} \left[\because e^{-2\infty}=0\right] =\frac{1}{10}$