$ \displaystyle (D^2y+3Dy+2y)=e^{-2t}.\sin t \\[2ex] $

$\displaystyle Taking\ laplace\ transform, \\[2ex] \displaystyle L\left[D^2y\right]+3L\left[Dy\right]+2L\left[y\right]=L\left[e^{-2t}.\sin t\right] \\[2ex] $

$\displaystyle Consider,\\[2ex] L[\sin {t]=\dfrac{1}{s^2+1}\\[2ex] L\left[e^{-2t}.\sin t\right]=\dfrac{1}{{(s+2)}^2+1}} \\[2ex] \displaystyle \therefore{}{[s}^2y-sy\left(0\right)-y^{'}(0)]+3[s\bar{y}-y(0)]+2\bar{y}=\frac{1}{{\left(s+2\right)}^2+1} \\[2ex] \displaystyle \therefore{}s^2y-0-1+3s\bar{y}-0+2\bar{y}=\frac{1}{s^2+2s+4+1} \\[2ex] \displaystyle \therefore{}\bar{y(}s^2+3s+2)=\frac{1}{s^2+2s+5}+1 \\[2ex] \displaystyle =\frac{1+s^2+2s+5}{s^2+2s+1}\\[2ex]=\ \dfrac{s^2+2s+6}{\left(s^2+2s+1\right)(s^2+3s+2)} \\[2ex] $

$\displaystyle y=L^{-1}\left[\frac{s^2+2s+6}{\left(s^2+2s+1\right)\left(s^2+3s+2\right)}\right] \\[2ex] $

$\displaystyle Consider,\\[2ex] L[\cos{at-\cos{bt}}]=L[\cos{at]-L[\cos{bt]}} \\[2ex] \displaystyle =\ \frac{s}{s^2+a^2}-\frac{s}{s^2+b^2} \\[2ex] \displaystyle \therefore{}\ L[\cos{at-\cos{bt}}]=\int_0^{\infty{}}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]ds \\[2ex] \displaystyle =\frac{1}{2}\int_0^{\infty{}}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]ds \\[2ex] \displaystyle ={\frac{1}{2}\displaystyle \left[ \log{\left(s^2+a^2\right)}-\log⁡(s^2+b^2)\right]}_s^{\infty{}} \\[2ex] \displaystyle ={\frac{1}{2}\displaystyle \left[\log{\left(\frac{s^2+a^2}{s^2+b^2}\right)}\right]}_s^{\infty{}} \\[2ex] \displaystyle =\frac{1}{2}\displaystyle\left[ 0-\log{\left(\frac{s^2+a^2}{s^2+b^2}\right)}\right] \\[2ex] \displaystyle \therefore{}\int_0^{\infty{}}e^{-st}\left(\frac{\cos{at-\cos{bt}}}{t}\right)dt=\frac{1}{2}\log {\left(\frac{s^2+b^2}{s^2+a^2}\right)} \\[2ex] $

$\displaystyle Put\ s=0, \\[2ex] \displaystyle \therefore{}\int_0^{\infty{}}\frac{\cos{at-\cos{bt}}}{t}dt\\[2ex] \displaystyle =\frac{1}{2}\log\left(\frac{0+b^2}{0+a^2}\right) \\[2ex] \displaystyle ={\log\left(\frac{b^2}{a^2}\right)}^{\frac{1}{2}} \\[2ex] \displaystyle =\log\left(\frac{b}{a}\right) \\[2ex] $