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

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

Answered by MathematicalUser2357 last updated on 10/Jan/25

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

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

$$I=\int \frac{\sin \left(\omega u\right)}{e^u}du$$$$I=-\frac{1}{\omega }\int \frac{1}{e^u}d\left(\cos \left(\omega u\right)\right)$$$$I=-\frac{1}{\omega }[\frac{\cos \left(\omega u\right)}{e^u}+\int \frac{\cos \left(\omega u\right)du}{e^u}]$$$$I=-\frac{1}{\omega }[\frac{\cos \left(\omega u\right)}{e^u}+\frac{1}{\omega }\int \frac{d(\sin \left(\omega u\right)}{e^u}]$$$$I=-\frac{1}{\omega }[\frac{\cos \left(\omega u\right)}{e^u}+\frac{1}{\omega }\times \frac{\sin \left(\omega u\right)}{e^u}+\frac{1}{\omega }\int \frac{\sin \left(\omega u\right)}{e^u}du]$$$$I=-\frac{1}{{\omega }^2}[\frac{\sin \left(\omega u\right)+\omega \cos \left(\omega u\right)}{e^u}+I]$$$$({\omega }^2+1)I=-\frac{\sin \left(\omega u\right)+\omega \cos \left(\omega u\right)}{e^u}$$$$\Rightarrow I=\int \frac{\sin \left(\omega u\right)}{e^u}du=-\frac{\sin \left(\omega u\right)+\omega \cos \left(\omega u\right)}{({\omega }^2+1)e^u}$$$$$$$$\int (e^{-\omega u}+\cos \left(u\right)-\frac{\sin \left(\omega u\right)}{e^u})du$$$$=\int e^{-\omega u}du+\int \cos \left(u\right)du-\int \frac{\sin \left(\omega u\right)}{e^u}du$$$$=-\frac{e^{-\omega u}}{\omega }+\sin u-\int \frac{\sin \left(\omega u\right)}{e^u}du$$$$=-\frac{e^{-\omega u}}{\omega }+\sin u+\frac{\sin \left(\omega u\right)+\omega \cos \left(\omega u\right)}{({\omega }^2+1)e^u}+C$$

Commented by alephnull last updated on 10/Jan/25

$$thank$$$$$$

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