Question

Find the \(L(\int_{0}^{t}sin⁡(u) cos⁡(2u)du)\).

(a) \(\frac{1}{2s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)

(b) \(\frac{1}{2s} \left [\frac{9}{s^2+9}-\frac{1}{s^2+1}\right ]\)

(c) \(\frac{1}{2s} \left [\frac{3}{s^2+9}+\frac{1}{s^2+1}\right ]\)

(d) \(\frac{1}{s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)

This question was posed to me in a job interview.

My question is taken from Table of General Properties of Laplace Transform in chapter Laplace Transform of Engineering Mathematics

Answer 1

The correct answer is (a) \(\frac{1}{2s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)

To explain: In the given question,

L(sin⁡(t) cos⁡(2t))=\(\frac{1}{2} L(sin⁡(3t)-sin⁡(t))\)

=\(\frac{1}{2} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)

By Laplace Transformation of an integral,

\(L(\int_{0}^{t}sin⁡(u) cos⁡(2u)du)=\frac{1}{2s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)