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Question Number 46542 by Umar last updated on 28/Oct/18

$$plshelp$$$$\mathit{F}\mathit{i}\mathit{n}\mathit{d}\mathit{L}\left({cos}^{2}t\right)$$

Commented by Umar last updated on 28/Oct/18

$$Laplacetransform$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$wehavegenerallyL\left(f\left(x\right)\right)={\int }_0^{\mathrm{\infty }}f\left(t\right)e^{-tx}dt\Rightarrow$$$$L\left({cos}^{2}x\right)={\int }_0^{\mathrm{\infty }}{cos}^{2}t{e}^{-tx}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}(1+cos\left(2t\right))e^{-tx}dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-tx}dt+\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-tx}cos\left(2t\right)dtbut$$$${\int }_0^{\mathrm{\infty }}{e}^{-tx}dt={[-\frac{1}{x}{e}^{-tx}]}_{t=0}^{\mathrm{\infty }}=\frac{1}{x}$$$${\int }_0^{\mathrm{\infty }}{e}^{-xt}cos\left(2t\right)dt=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-xt+i2t}dt\right)and$$$${\int }_0^{\mathrm{\infty }}{e}^{(-x+2i)t}dt={\left[\frac{1}{-x+2i}{e}^{(-x+2i)t}\right]}_{t=0}^{\mathrm{\infty }}=\frac{-1}{-x+2i}=\frac{1}{x-2i}=\frac{x+2i}{x^2+4}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{e}^{-xt}cos\left(2t\right)dt=\frac{x}{x^2+4}\Rightarrow L\left({cos}^{2}x\right)=\frac{1}{2x}+\frac{x}{2(x^2+4)}$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$anothermethodbyusinglinearityofLandtable$$$$L\left({cos}^{2}x\right)=L\left(\frac{1+cos\left(2x\right)}{2}\right)=\frac{1}{2}L\left(1\right)+\frac{1}{2}L\left(cos\left(2x\right)\right)$$$$=\frac{1}{2}\frac{1}{x}+\frac{1}{2}\frac{x}{x^2+4}=\frac{1}{2x}+\frac{x}{2x^2+8}.$$

Commented by Umar last updated on 28/Oct/18

$$thanks$$

Commented by maxmathsup by imad last updated on 28/Oct/18

$$youarewelcme.$$

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