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Question Number 46542 by Umar last updated on 28/Oct/18 | ||
$${pls}\:{help}\: \\ $$$$\:\boldsymbol{{Find}}\:\boldsymbol{{L}}\left({cos}^{\mathrm{2}} {t}\right) \\ $$ | ||
Commented by Umar last updated on 28/Oct/18 | ||
$${Laplace}\:{transform} \\ $$ | ||
Commented by maxmathsup by imad last updated on 28/Oct/18 | ||
$${we}\:{have}\:{generally}\:{L}\left({f}\left({x}\right)\right)=\int_{\mathrm{0}} ^{\infty} {f}\left({t}\right){e}^{−{tx}} {dt}\:\Rightarrow \\ $$$${L}\left({cos}^{\mathrm{2}} {x}\right)=\int_{\mathrm{0}} ^{\infty} \:{cos}^{\mathrm{2}} {t}\:{e}^{−{tx}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){e}^{−{tx}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:{but}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {dt}\:=\left[−\frac{\mathrm{1}}{{x}}{e}^{−{tx}} \right]_{{t}=\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{{x}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{xt}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}+{i}\mathrm{2}{t}} {dt}\right)\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−{x}+\mathrm{2}{i}\right){t}} {dt}\:=\left[\frac{\mathrm{1}}{−{x}+\mathrm{2}{i}}\:{e}^{\left(−{x}+\mathrm{2}{i}\right){t}} \right]_{{t}=\mathrm{0}} ^{\infty} \:=\frac{−\mathrm{1}}{−{x}+\mathrm{2}{i}}\:=\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}\:=\frac{{x}+\mathrm{2}{i}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow{L}\left({cos}^{\mathrm{2}} {x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\:+\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}\:} +\mathrm{4}\right)} \\ $$ | ||
Commented by maxmathsup by imad last updated on 28/Oct/18 | ||
$${another}\:{method}\:{by}\:{using}\:{linearity}\:{of}\:{L}\:{and}\:{table} \\ $$$${L}\left({cos}^{\mathrm{2}} {x}\right)={L}\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{L}\left(\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{L}\left({cos}\left(\mathrm{2}{x}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{{x}}{\mathrm{2}{x}^{\mathrm{2}\:} +\mathrm{8}}\:. \\ $$ | ||
Commented by Umar last updated on 28/Oct/18 | ||
$${thanks} \\ $$ | ||
Commented by maxmathsup by imad last updated on 28/Oct/18 | ||
$${you}\:{are}\:{welcme}. \\ $$ | ||