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Question Number 215498 by alephnull last updated on 08/Jan/25

∫(e^(−ωu) +cos(u)−((sin(ωu))/e^u ))du

$$\int\left({e}^{−\omega{u}} +\mathrm{cos}\left({u}\right)−\frac{\mathrm{sin}\left(\omega{u}\right)}{{e}^{{u}} }\right){du} \\ $$

Answered by MathematicalUser2357 last updated on 10/Jan/25

−(e^(−uω) /ω)+((e^(−u) sin uω)/((ω−i)(ω+i)))+((e^(−u) ωcos uω)/((ω−i)(ω+i)))+sin u+C e^(−u) (((sin uω)/(ω^2 +1))+((ωcos uω)/(ω^2 +1)))−(e^(−uω) /ω)+sin u+C ((e^(−u) sin uω)/(ω^2 +1))+((e^(−u) ωcos uω)/(ω^2 +1))−(e^(−uω) /ω)+sin u+C (((ω^2 +1)(ωsin u−e^(−uω) )+e^(−u) ω^2 cos uω+e^(−u) ωsin uω)/(ω(ω^2 +1)))+C The last two similar expressions are too long+C

$$−\frac{{e}^{−{u}\omega} }{\omega}+\frac{{e}^{−{u}} \mathrm{sin}\:{u}\omega}{\left(\omega−{i}\right)\left(\omega+{i}\right)}+\frac{{e}^{−{u}} \omega\mathrm{cos}\:{u}\omega}{\left(\omega−{i}\right)\left(\omega+{i}\right)}+\mathrm{sin}\:{u}+{C} \\ $$$${e}^{−{u}} \left(\frac{\mathrm{sin}\:{u}\omega}{\omega^{\mathrm{2}} +\mathrm{1}}+\frac{\omega\mathrm{cos}\:{u}\omega}{\omega^{\mathrm{2}} +\mathrm{1}}\right)−\frac{{e}^{−{u}\omega} }{\omega}+\mathrm{sin}\:{u}+{C} \\ $$$$\frac{{e}^{−{u}} \mathrm{sin}\:{u}\omega}{\omega^{\mathrm{2}} +\mathrm{1}}+\frac{{e}^{−{u}} \omega\mathrm{cos}\:{u}\omega}{\omega^{\mathrm{2}} +\mathrm{1}}−\frac{{e}^{−{u}\omega} }{\omega}+\mathrm{sin}\:{u}+{C} \\ $$$$\frac{\left(\omega^{\mathrm{2}} +\mathrm{1}\right)\left(\omega\mathrm{sin}\:{u}−{e}^{−{u}\omega} \right)+{e}^{−{u}} \omega^{\mathrm{2}} \mathrm{cos}\:{u}\omega+{e}^{−{u}} \omega\mathrm{sin}\:{u}\omega}{\omega\left(\omega^{\mathrm{2}} +\mathrm{1}\right)}+{C} \\ $$$$\mathrm{The}\:\mathrm{last}\:\mathrm{two}\:\mathrm{similar}\:\mathrm{expressions}\:\mathrm{are}\:\mathrm{too}\:\mathrm{long}+{C} \\ $$

Answered by mr W last updated on 10/Jan/25

I=∫((sin (ωu))/e^u )du I=−(1/ω)∫(1/e^u )d(cos (ωu)) I=−(1/ω)[((cos (ωu))/e^u )+∫((cos (ωu)du)/e^u )] I=−(1/ω)[((cos (ωu))/e^u )+(1/ω)∫((d(sin (ωu))/e^u )] I=−(1/ω)[((cos (ωu))/e^u )+(1/ω)×((sin (ωu))/e^u )+(1/ω)∫((sin (ωu))/e^u )du] I=−(1/ω^2 )[((sin (ωu)+ω cos (ωu))/e^u )+I] (ω^2 +1)I=−((sin (ωu)+ω cos (ωu))/e^u ) ⇒I=∫((sin (ωu))/e^u )du=−((sin (ωu)+ω cos (ωu))/((ω^2 +1)e^u )) ∫(e^(−ωu) +cos (u)−((sin (ωu))/e^u ))du =∫e^(−ωu) du+∫cos (u)du−∫((sin (ωu))/e^u )du =−(e^(−ωu) /ω)+sin u−∫((sin (ωu))/e^u )du =−(e^(−ωu) /ω)+sin u+((sin (ωu)+ω cos (ωu))/((ω^2 +1)e^u ))+C

$${I}=\int\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }{du} \\ $$$${I}=−\frac{\mathrm{1}}{\omega}\int\frac{\mathrm{1}}{{e}^{{u}} }{d}\left(\mathrm{cos}\:\left(\omega{u}\right)\right) \\ $$$${I}=−\frac{\mathrm{1}}{\omega}\left[\frac{\mathrm{cos}\:\left(\omega{u}\right)}{{e}^{{u}} }+\int\frac{\mathrm{cos}\:\left(\omega{u}\right){du}}{{e}^{{u}} }\right] \\ $$$${I}=−\frac{\mathrm{1}}{\omega}\left[\frac{\mathrm{cos}\:\left(\omega{u}\right)}{{e}^{{u}} }+\frac{\mathrm{1}}{\omega}\int\frac{{d}\left(\mathrm{sin}\:\left(\omega{u}\right)\right.}{{e}^{{u}} }\right] \\ $$$${I}=−\frac{\mathrm{1}}{\omega}\left[\frac{\mathrm{cos}\:\left(\omega{u}\right)}{{e}^{{u}} }+\frac{\mathrm{1}}{\omega}×\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }+\frac{\mathrm{1}}{\omega}\int\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }{du}\right] \\ $$$${I}=−\frac{\mathrm{1}}{\omega^{\mathrm{2}} }\left[\frac{\mathrm{sin}\:\left(\omega{u}\right)+\omega\:\mathrm{cos}\:\left(\omega{u}\right)}{{e}^{{u}} }+{I}\right] \\ $$$$\left(\omega^{\mathrm{2}} +\mathrm{1}\right){I}=−\frac{\mathrm{sin}\:\left(\omega{u}\right)+\omega\:\mathrm{cos}\:\left(\omega{u}\right)}{{e}^{{u}} } \\ $$$$\Rightarrow{I}=\int\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }{du}=−\frac{\mathrm{sin}\:\left(\omega{u}\right)+\omega\:\mathrm{cos}\:\left(\omega{u}\right)}{\left(\omega^{\mathrm{2}} +\mathrm{1}\right){e}^{{u}} } \\ $$$$ \\ $$$$\int\left({e}^{−\omega{u}} +\mathrm{cos}\:\left({u}\right)−\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }\right){du} \\ $$$$=\int{e}^{−\omega{u}} {du}+\int\mathrm{cos}\:\left({u}\right){du}−\int\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }{du} \\ $$$$=−\frac{{e}^{−\omega{u}} }{\omega}+\mathrm{sin}\:{u}−\int\frac{\mathrm{sin}\:\left(\omega{u}\right)}{{e}^{{u}} }{du} \\ $$$$=−\frac{{e}^{−\omega{u}} }{\omega}+\mathrm{sin}\:{u}+\frac{\mathrm{sin}\:\left(\omega{u}\right)+\omega\:\mathrm{cos}\:\left(\omega{u}\right)}{\left(\omega^{\mathrm{2}} +\mathrm{1}\right){e}^{{u}} }+{C} \\ $$

Commented by alephnull last updated on 10/Jan/25

thank

$${thank} \\ $$$$ \\ $$

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