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

pls help    Find L(cos^2 t)

$${pls}\:{help}\: \\ $$$$\:\boldsymbol{{Find}}\:\boldsymbol{{L}}\left({cos}^{\mathrm{2}} {t}\right) \\ $$

Commented by Umar last updated on 28/Oct/18

Laplace transform

$${Laplace}\:{transform} \\ $$

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

we have generally L(f(x))=∫_0 ^∞ f(t)e^(−tx) dt ⇒  L(cos^2 x)=∫_0 ^∞  cos^2 t e^(−tx) dt=(1/2)∫_0 ^∞ (1+cos(2t))e^(−tx) dt  =(1/2)∫_0 ^∞ e^(−tx) dt +(1/2)∫_0 ^∞  e^(−tx)  cos(2t)dt but   ∫_0 ^∞  e^(−tx) dt =[−(1/x)e^(−tx) ]_(t=0) ^∞ =(1/x)  ∫_0 ^∞   e^(−xt)  cos(2t)dt =Re(∫_0 ^∞  e^(−xt+i2t) dt) and  ∫_0 ^∞   e^((−x+2i)t) dt =[(1/(−x+2i)) e^((−x+2i)t) ]_(t=0) ^∞  =((−1)/(−x+2i)) =(1/(x−2i)) =((x+2i)/(x^2  +4)) ⇒  ∫_0 ^∞  e^(−xt)  cos(2t)dt = (x/(x^2  +4)) ⇒L(cos^2 x)=(1/(2x))  +(x/(2(x^(2 ) +4)))

$${we}\:{have}\:{generally}\:{L}\left({f}\left({x}\right)\right)=\int_{\mathrm{0}} ^{\infty} {f}\left({t}\right){e}^{−{tx}} {dt}\:\Rightarrow \\ $$$${L}\left({cos}^{\mathrm{2}} {x}\right)=\int_{\mathrm{0}} ^{\infty} \:{cos}^{\mathrm{2}} {t}\:{e}^{−{tx}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){e}^{−{tx}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:{but}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {dt}\:=\left[−\frac{\mathrm{1}}{{x}}{e}^{−{tx}} \right]_{{t}=\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{{x}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{xt}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}+{i}\mathrm{2}{t}} {dt}\right)\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−{x}+\mathrm{2}{i}\right){t}} {dt}\:=\left[\frac{\mathrm{1}}{−{x}+\mathrm{2}{i}}\:{e}^{\left(−{x}+\mathrm{2}{i}\right){t}} \right]_{{t}=\mathrm{0}} ^{\infty} \:=\frac{−\mathrm{1}}{−{x}+\mathrm{2}{i}}\:=\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}\:=\frac{{x}+\mathrm{2}{i}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow{L}\left({cos}^{\mathrm{2}} {x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\:\:+\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}\:} +\mathrm{4}\right)} \\ $$

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

another method by using linearity of L and table  L(cos^2 x)=L(((1+cos(2x))/2))=(1/2)L(1) +(1/2)L(cos(2x))  =(1/2) (1/x) +(1/2) (x/(x^2  +4)) =(1/(2x)) +(x/(2x^(2 ) +8)) .

$${another}\:{method}\:{by}\:{using}\:{linearity}\:{of}\:{L}\:{and}\:{table} \\ $$$${L}\left({cos}^{\mathrm{2}} {x}\right)={L}\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{L}\left(\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{L}\left({cos}\left(\mathrm{2}{x}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{{x}}{\mathrm{2}{x}^{\mathrm{2}\:} +\mathrm{8}}\:. \\ $$

Commented by Umar last updated on 28/Oct/18

thanks

$${thanks} \\ $$

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

you are welcme.

$${you}\:{are}\:{welcme}. \\ $$

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