Question

The Laplace transform of $e^{-4t}\frac{\sin 3t}{t}$ is

• A. ${\tan }^{-1}\frac{3}{s+4}$
• B. ${\tan }^{-1}\frac{3}{s+4}$
• C. ${\tan }^{-1}\frac{s+4}{3}$
• D. ${\cot }^{-1}\frac{s}{3}$

Answer 1

The correct option is B ${\tan }^{-1}\frac{3}{s+4}$

$L\left\{\sin 3t\right\}=\frac{3}{s^2+3^2}$

$L\left\{\frac{\sin 3t}{t}\right\}={\int }_s^{\infty }\frac{3}{s^2+9}ds$

$={\left[\frac{3}{3}{\tan }^{-1}\frac{s}{3}\right]}_s^{\infty }$

$=\frac{\pi }{2}-{\tan }^{-1}\frac{s}{3}={\cot }^{-1}\frac{s}{3}$

$L=\left[e^{-4t}\frac{\sin 3t}{t}\right]={\cot }^{-1}\frac{s+4}{3}$

$={\tan }^{-1}\frac{3}{s+4}$