Question Number 228113 by TonyCWX last updated on 25/Mar/26
![Just want to share my work here. 1. ∫ e^x sin(ax) dx Let I = ∫ e^x sin(ax) dx =^(IBP) e^x sin(ax) − a∫ e^x cos(ax) dx =^(IBP) e^x sin(ax) − a[e^x cos(ax) + a∫ e^x sin(ax) dx] = e^x sin(ax) − ae^x cos(ax) − a^2 I I = e^x sin(ax) − ae^x cos(ax) − a^2 I (a^2 + 1)I = e^x sin(ax) − ae^x cos(ax) I = ((e^x sin(ax))/(a^2 + 1)) − ((ae^x cos(ax))/(a^2 + 1)) + C determinant (((∫ e^x sin(ax) dx = ((e^x sin(ax))/(a^2 + 1)) − ((ae^x cos(ax))/(a^2 + 1)) + C))) 2. ∫ e^x cos(ax) dx Let I = ∫ e^x cos(ax) dx =^(IBP) e^x cos(ax) + a∫ e^x sin(ax) dx =^(IBP) e^x cos(ax) + a[e^x sin(ax) − a∫ e^x cos(ax) dx] = e^x cos(ax) + ae^x sin(ax) − a^2 I I = e^x cos(ax) + ae^x sin(ax) − a^2 I (a^2 + 1)I = e^x cos(ax) + ae^x sin(ax) I = ((e^x cos(ax))/(a^2 + 1)) + ((ae^x sin(ax))/(a^2 + 1)) + C determinant (((∫ e^x cos(ax) dx = ((e^x cos(ax))/(a^2 + 1)) + ((ae^x sin(ax))/(a^2 + 1)) + C))) Note: C is an arbitrary constant Open for corrections.](https://www.tinkutara.com/question/Q228113.png)
$$\mathrm{Just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{share}\:\mathrm{my}\:\mathrm{work}\:\mathrm{here}.\: \\ $$$$\mathrm{1}.\:\int\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:{dx} \\ $$$$\mathrm{Let}\:{I}\:=\:\int\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:{dx} \\ $$$$\overset{\mathrm{IBP}} {=}\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{a}\int\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:{dx} \\ $$$$\overset{\mathrm{IBP}} {=}\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{a}\left[{e}^{{x}} \mathrm{cos}\left({ax}\right)\:+\:{a}\int\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:{dx}\right] \\ $$$$=\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{ae}^{{x}} \mathrm{cos}\left({ax}\right)\:−\:{a}^{\mathrm{2}} {I} \\ $$$$ \\ $$$${I}\:=\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{ae}^{{x}} \mathrm{cos}\left({ax}\right)\:−\:{a}^{\mathrm{2}} {I} \\ $$$$\left({a}^{\mathrm{2}} \:+\:\mathrm{1}\right){I}\:=\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{ae}^{{x}} \mathrm{cos}\left({ax}\right) \\ $$$${I}\:=\:\frac{{e}^{{x}} \mathrm{sin}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\frac{{ae}^{{x}} \mathrm{cos}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:{C} \\ $$$$ \\ $$$$\begin{array}{|c|}{\int\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:{dx}\:=\:\frac{{e}^{{x}} \mathrm{sin}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\frac{{ae}^{{x}} \mathrm{cos}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:{C}}\\\hline\end{array} \\ $$$$ \\ $$$$\mathrm{2}.\:\int\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:{dx} \\ $$$$\mathrm{Let}\:{I}\:=\:\int\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:{dx} \\ $$$$\overset{\mathrm{IBP}} {=}{e}^{{x}} \mathrm{cos}\left({ax}\right)\:+\:{a}\int\:{e}^{{x}} \mathrm{sin}\left({ax}\right)\:{dx} \\ $$$$\overset{\mathrm{IBP}} {=}{e}^{{x}} \mathrm{cos}\left({ax}\right)\:+\:{a}\left[{e}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{a}\int\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:{dx}\right] \\ $$$$=\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:+\:{ae}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{a}^{\mathrm{2}} {I} \\ $$$$ \\ $$$${I}\:=\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:+\:{ae}^{{x}} \mathrm{sin}\left({ax}\right)\:−\:{a}^{\mathrm{2}} {I} \\ $$$$\left({a}^{\mathrm{2}} \:+\:\mathrm{1}\right){I}\:=\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:+\:{ae}^{{x}} \mathrm{sin}\left({ax}\right) \\ $$$${I}\:=\:\frac{{e}^{{x}} \mathrm{cos}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:\frac{{ae}^{{x}} \mathrm{sin}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:{C} \\ $$$$ \\ $$$$\begin{array}{|c|}{\int\:{e}^{{x}} \mathrm{cos}\left({ax}\right)\:{dx}\:=\:\frac{{e}^{{x}} \mathrm{cos}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:\frac{{ae}^{{x}} \mathrm{sin}\left({ax}\right)}{{a}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:{C}}\\\hline\end{array} \\ $$$$ \\ $$$$\mathrm{Note}:\:{C}\:\mathrm{is}\:\mathrm{an}\:\mathrm{arbitrary}\:\mathrm{constant} \\ $$$$\mathrm{Open}\:\mathrm{for}\:\mathrm{corrections}. \\ $$
Commented by AgniMath last updated on 26/Mar/26
Commented by Abod last updated on 28/Mar/26
$${i}\:{am}\:{a}\:{child}\:{but}\:{what}\:{is}\:{that}\:\left\{{sorry}\:{i}\:{can}.{t}\:{spaeck}\:{englech}\right\}{o} \\ $$
Commented by TonyCWX last updated on 29/Mar/26
$$\mathrm{Bernoulli}'\mathrm{s}\:\mathrm{Rule}: \\ $$$$\int\:{u}\:{dv}\:=\:{uv}_{\mathrm{1}} \:−\:{u}'{v}_{\mathrm{2}} \:+\:{u}''{v}_{\mathrm{3}} \:−\:{u}'''{v}_{\mathrm{4}} \:+\:… \\ $$$$\mathrm{Where} \\ $$$$\:{u}^{{n}} \:=\:{n}^{\mathrm{th}} \:\mathrm{derivative}\:\mathrm{of}\:{u}\: \\ $$$${v}_{\mathrm{1}} \:=\:\int\:{v}\:{dv} \\ $$$${v}_{\mathrm{2},\mathrm{3},\mathrm{4},…} \:=\:\mathrm{successive}\:\mathrm{integral}\:\mathrm{of}\:{v}_{\mathrm{1}} \\ $$