Question Number 30476 by abdo imad last updated on 22/Feb/18
$${find}\:{L}\left({cos}^{\mathrm{2}} {x}\right)\:{and}\:{L}\left({sin}^{\mathrm{2}} {x}\right)\:{L}\:{is}\:{laplace}\:{transform}. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
$${L}\left({cos}^{\mathrm{2}} {x}\right)={L}\left(\frac{{cos}\left(\mathrm{2}{x}\right)+\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({L}\left({cos}\left(\mathrm{2}{x}\right)\right)+{L}\left(\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{{s}}\right) \\ $$$${samething}\:{for}\:\:{L}\left({sin}^{\mathrm{2}} \left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{s}}−\frac{\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\right) \\ $$
Commented by sma3l2996 last updated on 24/Feb/18
$${L}\left({sin}^{\mathrm{2}} \left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{s}}−\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\right) \\ $$