📂Odinary Differential Equations

Laplace Transform of Trigonometric Functions

Formulas¹

The Laplace transforms of sine and cosine are as follows.

$$\mathcal{L} \left\{ \sin (at) \right\} = \dfrac{a}{s^2+a^2},\quad s>0$$

$$\mathcal{L} \left\{ \cos (at) \right\} = \dfrac{s}{s^2+a^2},\quad s>0$$

Derivation

$\sin (at)$

$$\begin{align*} \mathcal{L} \left\{ \sin (at) \right\}& =\displaystyle \int_{0}^\infty e^{-st}\sin(at)dt \\ &= \lim \limits_{A \to \infty} \left[-\dfrac{1}{a}e^{-st}\cos (at) \right]_{0}^A+ \lim \limits_{A \to \infty} \int _{0}^\infty -\dfrac{s}{a}e^{-st} \cos (at)dt \\ &= \dfrac{1}{a} - \lim \limits_{A \to \infty} \dfrac{s}{a} \left[ \dfrac{1}{a} \left[ e^{-st}\sin (at) \right]_{0}^A + \dfrac{s}{a}\int _{0}^A e^{-st} \sin (at) dt \right] \\ &=\dfrac{1}{a} - \dfrac{s^2}{a^2} \int_{0}^\infty e^{-st} \sin (at) dt \end{align*}$$

Since $\mathcal{L} \left\{ \sin (at) \right\} = \displaystyle \int_{0}^\infty e^{-st} \sin (at) dt$ holds,

$$\begin{align*} \implies& &\dfrac{a^2+s^2}{a^2} \int _{0}^\infty e^{-st} \sin (at) dt &= \dfrac{1}{a} \\ \implies& &\int_{0}^\infty e^{-st} \sin (at)dt &=\dfrac{a}{s^2+a^2} \end{align*}$$

provided that $\lim \limits_{A \to \infty} e^{-sA}\sin (aA)=0$ is satisfied, hence $s>0$

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$\cos (at)$

Using the result of $\sin$, the Laplace transform of $\cos$ can be obtained much more easily and briefly.

$$\begin{align*} \mathcal{ L } \left\{ \cos (at) \right\} &=\int _{0}^\infty e^{-st} \cos (at) dt \\ &= \lim \limits_{A \to \infty} \dfrac{1}{a} \left[ e^{-st} \sin (at) \right]_{0}^A + \dfrac{s}{a} \int_{0}^\infty e^{-st} \sin (at) dt \\ &= \dfrac{s}{a} \dfrac{a}{s^2+a^2} \\ &=\dfrac{s}{s^2+a^2} \end{align*}$$

provided that $s>0$

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See Also

• Laplace transform table