*$i) \ \ \ Consider,\\[2ex] L\left\{\cos{bt-\cos{at}}\right\}\\[2ex] \therefore{}L\left\{\cos{bt}\right\}-L\left\{\cos{at}\right\}=\frac{s}{s^2+b^2}-\frac{s}{s^2+a^2}\\[2ex] \displaystyle \therefore{}L\left[\frac{1}{t}\left(\cos{bt-\cos{at}}\right)\right]=\int_s^{\infty{}}\left(\frac{s}{s^2+b^2}-\frac{s}{s^2+a^2}\right)ds\ \\[2ex] \displaystyle =\frac{1}{2}\int_s^{\infty{}}\left(\frac{2s}{s^2+b^2}-\frac{2s}{s^2+a^2}\right)ds\\[2ex] \displaystyle =\frac{1}{2}{\left[\log\left(s^2+b^2\right)-\log\left(s^2+a^2\right)\right]}_s^{\infty{}}\\[2ex] \displaystyle =\frac{1}{2}{\left[\log\left(\frac{s^2+b^2}{s^2+a^2}\right)\right]}_s^{\infty{}}\\[2ex] \displaystyle =\frac{1}{2}\left[0-\log\left(\frac{s^2+b^2}{s^2+a^2}\right)\right]\\[2ex] \displaystyle =\frac{1}{2}\left[\log\left(\frac{s^2+b^2}{s^2+a^2}\right)\right]\\$*** $ii)\ Consider,\[2ex] L\left{\sin t\right}=\frac{1}{s^2+1}\[2ex] \displaystyle \therefore{}L\left{\frac{\sin t}{t}\right}=\int_s^{\infty{}}\left(\frac{1}{s^2+1}\right)ds\ \[2ex] \displaystyle =\frac{1}{2}\int_s^{\infty{}}\left(\frac{2}{s^2+1}\right)ds$ $ \displaystyle =\frac{1}{2}{\left[\log\left(s^2+1\right)\right]}_s^{\infty{}}\[2ex]

\displaystyle =\frac{1}{2}\left[0-\log\left(s^2+1\right)\right]\[2ex] \displaystyle =\frac{1}{2}\left[\log\left(s^2+1\right)\right]\[2ex] $ $ Now , f(0)= {\log}_{t=0}{\frac{\sin t}{t}=0}\[3ex] Using\ Theorem,\[2ex] \displaystyle L\left{f^{'}(t)\right}=-f(0)+sL\left{f(t)\right}L\left[\frac{d}{dt}\left(\frac{\sin t}{t}\right)\right]\[2ex] \displaystyle =0+s\bullet{}\frac{1}{2}\left[\log\left(s^2+1\right)\right]\ \[2ex] \displaystyle =\ \frac{s}{2}\left[\log\left(s^2+1\right)\right] $