Topic: Engineering Mathematics
Laplace transform of $sin^{3}2t$ is
- A. $\frac{24}{(s^{2}+4)(s^{2}+36)}$
- B. $\frac{1}{(s^{2}+4)(s^{2}+64)}$
- C. $\frac{48}{(s^{2}+4)(s^{2}+36)}$
- D. $\frac{64}{(s^{2}+4)(s^{2}+36)}$
Answer: C. $\frac{48}{(s^{2}+4)(s^{2}+36)}$
Explanation: We know that, $sin~3x=3~sin~x-4~sin^{3}x \implies sin^{3}x=\frac{3~sin~x-sin~3x}{4}$. Substituting $x=2t$, we get $sin^{3}2t=\frac{3~sin~2t-sin~6t}{4}$. The Laplace transform for the above is $\frac{1}{4}[\frac{3\times2}{s^{2}+2^{2}}-\frac{6}{s^{2}+6^{2}}] = \frac{1}{4}(\frac{6}{s^{2}+4}-\frac{6}{s^{2}+36}) = \frac{48}{(s^{2}+4)(s^{2}+36)}$.