$\displaystyle Consider\\[2ex] \displaystyle L[\cos hu]=\ \frac{s}{s^2-1^2} \\[2ex] \displaystyle \therefore{}\ L\{e^u\cos h{u\}=\frac{s-1}{({s-1)}^2-1}}….(first\ shifting\ property) \\[2ex] \displaystyle =\frac{s-1}{({s)}^2-2s} \\[2ex] $

$\displaystyle \therefore{}L\left[\int_0^te^u\cos hu\ du\right ]=\frac{1}{s}.\left\{\frac{s-1}{s\left(s-2\right)}\right\}=\frac{s-1}{s^2(s-2)} \\[2ex] \displaystyle \therefore{}L\left[e^{-t}\int_0^te^u\cos hu\ du\right]=\frac{(s+1)-1}{({s+1)}^2[(s+1)-2]}….(first\ shifting\ property) \\[2ex] \displaystyle =\frac{s}{{\left(s+1\right)}^2(s-1)} \\[2ex] $

$\displaystyle Consider\\[2ex] \displaystyle \int_0^{2\pi{}}\frac{e^{2i\theta{}}}{(5+4\cos{\theta{})}}d\theta{}\ \ \ \\[2ex] \displaystyle Put\ z=\ e^{i\theta{}}\ \ \therefore{}dz=ie^{i\theta{}}.d\theta{}. \\[2ex] \displaystyle \therefore{}\ dz=\ izd\theta{}\ \ \ \therefore{}\ d\theta{}=\frac{dz}{iz} \\[2ex] $

$\displaystyle And\ \cos{\theta{}}=\ \frac{e^{i\theta{}+\ }e^{-i\theta{}}}{2}\ =\ \frac{z+(\frac{1}{z})}{2} \\[2ex] \displaystyle \therefore{}\int_0^{2\pi{}}\frac{e^{2i\theta{}}}{(5+4\cos{\theta{}}}d\theta{}\ \ \ =\int_c\frac{z^2}{5+4\left(\frac{z+\left(\frac{1}{z}\right)}{2}\right)}.\frac{dz}{iz} \\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_c\frac{z^2}{i(2z^2+5z+2)}dz \\[2ex] $

$\displaystyle where\ C\ is\ the\ circle\ \left\vert{}z\right\vert{}=1. \\[2ex] \displaystyle Now\ the\ poles\ are\ given\ by\ {2z}^2+5z+2=0\\[2ex] \therefore{}\ (2z+1)(z+2)=0 \\[2ex] \displaystyle \therefore{}\ z\ =\ -1/2\ and\ z\ =\ -2 \\[2ex] $

$\displaystyle The\ pole\ z\ =\ -1/2\ lies\ inside\ the\ unit\ circle\ and\ the\ pole\ z=\ -2\ lies\ outside. \\[2ex] $

$\displaystyle Now,\ Residue\ of\ f(z)\ (at\ z=\ -1/2) \\[2ex] \displaystyle =\ \lim_{z\rightarrow{}-\frac{1}{2}}\ \left(z+\frac{1}{2}\right)\bullet{}{\frac{z^2}{2\left[z+\left(\frac{1}{2}\right)\right]\left(z+2\right)i}} \\[2ex] \displaystyle =\frac{{(-\frac{1}{2})}^2}{2\left[\left(-\frac{1}{2}\right)+2\right]i}\\[2ex] \displaystyle =\ \frac{1}{12i} \\[2ex] $

$\displaystyle \therefore{}\int_0^{2\pi{}}\frac{e^{2i\theta{}}}{(5+4\cos{\theta{})}}d\theta{}\ \ \ =\ 2\pi{}i(\frac{1}{12i})\ =\ \frac{\pi{}}{6} \\[2ex] \displaystyle \int_0^{2\pi{}}\frac{\cos{2\theta{}}}{(5+4\cos{\theta{})}}\ d\theta{}\ =\ Real\ part\ of\ \int_0^{2\pi{}}\frac{e^{2i\theta{}}}{(5+4\cos{\theta{})}}d\theta{}\ \ \ =\ \frac{\pi{}}{6} \\[2ex]$