$Consider\ L\left[\sin{\left(t+\alpha{}\right)}\cos{\left(t-\alpha{}\right)}\right]\\[2ex]=L\left[\frac{1}{2}\ (\sin{2t}+\sin{2\alpha{})}\right] \\[2ex] \displaystyle =\frac{1}{2}L\left[\sin{2t\ }\right]+\frac{1}{2}L\left[\sin{2\alpha{}}\right]=\frac{1}{2}\frac{2}{s^2+2^2}+\frac{1}{2}\frac{\sin{2\alpha{}}}{s}\ \ \\[2ex] \displaystyle =\frac{1}{s^2+4\ }+\frac{\sin{2\alpha{}\ }}{2s} \\[2ex] $

$From \ the \ definition \ of \ Laplace \ Transform, \\[3ex] \displaystyle \int_0^{\infty{}}e^{-st}\ \sin{\left(t+\alpha{}\right)}\cos{(t-\alpha{})}dt=\ \frac{1}{s^2+4\ }+\frac{\sin{2\alpha{}\ }}{2s} \\[5ex] Put s = 2, \\[3ex] \displaystyle \int_0^{\infty{}}e^{-2t}\ \sin{\left(t+\alpha{}\right)}\cos{(t-\alpha{})}dt=\ \frac{1}{2^2+4\ }+\frac{\sin{2\alpha{}\ }}{2\times{}2} \\[2ex] \displaystyle \frac{3}{8}=\frac{1}{4+4}+\frac{\sin{2\alpha{}}}{4}\ \\[2ex] \displaystyle \sin{2\alpha{}}=4\left(\frac{3-1}{8}\right)=1\ \\[2ex] \displaystyle 2\alpha{}=\frac{\pi{}}{2}+2\pi{}n\ \\[2ex] \displaystyle \alpha{}=\frac{\pi{}}{4}+\pi{}n\ \ \ \ \ \ (n\in{}I) \\[2ex] $