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Question Number 5629 by LMTV last updated on 23/May/16

$$\mathcal{L}\left(\sin {}^{2}t\right)=?$$

Commented by FilupSmith last updated on 23/May/16

$$\mathrm{Laplace}\mathrm{transform}?$$

Commented by LMTV last updated on 23/May/16

$$yes>⌣<$$

Commented by FilupSmith last updated on 23/May/16

$$\mathcal{L}\left({\sin }^{2}t\right)={\int }_0^{\mathrm{\infty }}{e}^{-st}{\sin }^2\left(t\right)dt$$$${\sin }^{2}t=\frac{1}{2}(1-\cos \left(2t\right))$$$$\mathcal{L}\left({\sin }^{2}t\right)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-st}(1-\cos \left(2t\right))dt$$$$\mathrm{unsure}\mathrm{how}\mathrm{to}\mathrm{solve}$$

Commented by Yozzii last updated on 23/May/16

$$e^{(a+bi)x}=e^{ax}{e}^{bix}=e^{ax}(cosbx+isinbx)$$$$\Rightarrow Re\int e^{(a+bi)x}dx=\int e^{ax}cosbxdx$$$$\therefore \int e^{ax}cosbxdx=Re\left(\frac{1}{a+bi}{e}^{(a+bi)x}\right)+C$$$$\int e^{ax}cosbxdx=Re\left(\frac{(a-bi)e^{(a+bi)x}}{a^2+b^2}\right)+C$$$$\int e^{ax}cosbxdx=\frac{e^{ax}(acosbx+bsinbx)}{a^2+b^2}+C$$$$Leta=-sandb=2.$$$$\therefore \int e^{-sx}cos2xdx=\frac{e^{-sx}(2sin2x-scos2x)}{s^2+4}+C$$$$\therefore \mathcal{L}\left({sin}^{2}t\right)=\frac{1}{2}\underset{m\to \mathrm{\infty }}{\lim }(\frac{-e^{-st}}{s}+\frac{e^{-st}(scos2t-2sin2t)}{s^2+4}){\mid }_0^m$$$$\mathcal{L}\left({sin}^{2}t\right)=0.5\underset{m\to \mathrm{\infty }}{\lim }(e^{-sm}(\frac{scos2m-2sin2m}{s^2+4}-\frac{1}{s})+\frac{1}{s}-\frac{s}{s^2+4})$$$$\mathcal{L}\left({sin}^{2}t\right)=0.5\left(\frac{s^2+4-s^2}{s(s^2+4)}\right)$$$$\mathcal{L}\left({sin}^{2}t\right)=\frac{2}{s(s^2+4)}$$

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