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Evaluate-L-tan-1-t-1-t-solution-F-s-L-tan-1-t-1-t-sF-s-pi-2-L-t-2-1-t-4-t-2-1-s-sF-s-pi-2-L-1-2-t-2-3-t-1-1-2-t-2-3-t-1-s-s

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Question Number 219589 by Nicholas666 last updated on 29/Apr/25
Evaluate; L(tan^(−1) (t−(1/t))) solution; ⇒F(s)= L(tan^(−1) (t−(1/t))) ⇔ sF(s)+(π/2)=L(((t^2 +1)/(t^4 −t^2 +1)))(s) ⇒ sF(s)+(π/2)=L(((1/2)/(t^2 −(√3) t +1))+((1/2)/(t^2 −(√3) t+1)))(s) ⇒ sF(s)+(π/2)=(1/2)L((1/((t−((√3)/2))^2 +(1/4))))(s)+(1/2)L((1/((t+((√3)/2))^2 +(1/4))))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s)+(1/2)e^(((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) (1/(1/2)) sin ((1/2) s)∗(1/s)+(1/2)e^(((√3)/2) s) (1/(1/2))sin((1/2) s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) L(sin((1/2) t))(s)∗(1/s)+e^(((√3)/2) s) L(sin((1/2) t))(s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/(2 ))s) ((1/2)/(s^2 +1/4))∗(1/s) Final Answer; F(s)=(1/s)(e^(−((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s))−(π/(2s))
$${Evaluate};\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{solution}; \\ $$$$\:\Rightarrow{F}\left({s}\right)=\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\Leftrightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\:\:\:\:\: \\ $$$$\mathrm{Final}\:\mathrm{Answer}; \\ $$$$\:\:\:{F}\left({s}\right)=\frac{\mathrm{1}}{{s}}\left({e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\right)−\frac{\pi}{\mathrm{2}{s}}\:\: \\ $$$$ \\ $$
Commented by Nicholas666 last updated on 29/Apr/25
hello friends.. is there a better solution?
$${hello}\:{friends}.. \\ $$$${is}\:{there}\:{a}\:{better}\:{solution}? \\ $$
Commented by Nicholas666 last updated on 29/Apr/25
please show me the corret solution guys, I am doing a mistake here. ????????????????????
$${please}\:{show}\:{me}\:{the}\:{corret}\:{solution}\:{guys}, \\ $$$${I}\:{am}\:{doing}\:{a}\:{mistake}\:{here}. \\ $$$$???????????????????? \\ $$$$ \\ $$

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