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Question Number 115743 by bemath last updated on 28/Sep/20
∫ e^(ax) .sin bx dx =? by complex number
$$\int\:{e}^{{ax}} .\mathrm{sin}\:{bx}\:{dx}\:=? \\ $$$${by}\:{complex}\:{number} \\ $$
Answered by Ar Brandon last updated on 28/Sep/20
I=∫e^(ax) sinbxdx=(1/(2i))∫e^(ax) ∙(e^(bxi) −e^(−bxi) )dx =(1/(2i))∫(e^((a+bi)x) −e^((a−bi)x) )dx =(1/(2i)){(e^((a+bi)x) /(a+bi))−(e^((a−bi)x) /(a−bi))}+C
$$\mathcal{I}=\int\mathrm{e}^{\mathrm{a}{x}} \mathrm{sinb}{x}\mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{2}{i}}\int\mathrm{e}^{\mathrm{a}{x}} \centerdot\left(\mathrm{e}^{\mathrm{b}{xi}} −\mathrm{e}^{−\mathrm{b}{xi}} \right)\mathrm{d}{x} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\int\left(\mathrm{e}^{\left(\mathrm{a}+\mathrm{b}{i}\right){x}} −\mathrm{e}^{\left(\mathrm{a}−\mathrm{b}{i}\right){x}} \right)\mathrm{d}{x} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\frac{\mathrm{e}^{\left(\mathrm{a}+\mathrm{b}{i}\right){x}} }{\mathrm{a}+\mathrm{b}{i}}−\frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{b}{i}\right){x}} }{\mathrm{a}−\mathrm{b}{i}}\right\}+\mathrm{C} \\ $$
Commented by bemath last updated on 28/Sep/20
santuy bro. gave kudos
$${santuy}\:{bro}.\:{gave}\:{kudos} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Sep/20
∫e^(ax) .sinbx dx =(1/(2i))∫e^(ax) (e^(bxi) −e^(−bxi) ) =(1/(2i))∫e^(x(a+bi)) −(1/(2i))∫e^(x(a−bi)) dx =(e^(x(a+bi)) /(2i(a+bi)))−(e^(x(a−bi)) /(2i(a−bi))) =(e^(xz) /(2iz))−(e^(xz^− ) /(2iz^− ))+C (z=a+bi z^− =a−bi)
$$\int\mathrm{e}^{\mathrm{ax}} .\mathrm{sinbx}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\int\mathrm{e}^{\mathrm{ax}} \left(\mathrm{e}^{\mathrm{bxi}} −\mathrm{e}^{−\mathrm{bxi}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\int\mathrm{e}^{\mathrm{x}\left(\mathrm{a}+\mathrm{bi}\right)} −\frac{\mathrm{1}}{\mathrm{2i}}\int\mathrm{e}^{\mathrm{x}\left(\mathrm{a}−\mathrm{bi}\right)} \mathrm{dx}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\frac{\mathrm{e}^{\mathrm{x}\left(\mathrm{a}+\mathrm{bi}\right)} }{\mathrm{2i}\left(\mathrm{a}+\mathrm{bi}\right)}−\frac{\mathrm{e}^{\mathrm{x}\left(\mathrm{a}−\mathrm{bi}\right)} }{\mathrm{2i}\left(\mathrm{a}−\mathrm{bi}\right)} \\ $$$$=\frac{\mathrm{e}^{\mathrm{xz}} }{\mathrm{2iz}}−\frac{\mathrm{e}^{\mathrm{x}\overset{−} {\mathrm{z}}} }{\mathrm{2i}\overset{−} {\mathrm{z}}}+\mathrm{C}\:\:\:\:\:\:\:\:\left(\mathrm{z}=\mathrm{a}+\mathrm{bi}\:\:\overset{−} {\mathrm{z}}=\mathrm{a}−\mathrm{bi}\right) \\ $$
Answered by mathmax by abdo last updated on 28/Sep/20
∫ e^(ax) sin(bx)dx =Im(∫ e^(ax+ibx) dx) and ∫ e^((a+ib)x) dx =(1/(a+ib)) e^((a+ib)x) +c =((a−ib)/(a^2 +b^2 )) .e^(ax) {cos(bx)+isin(bx)} +c =(e^(ax) /(a^2 +b^2 )){acos(bx)+iasin(bx)−ibcos(bx) +bsin(bx)} ⇒ ∫ e^(ax) sin(bx)dx =(e^(ax) /(a^2 +b^2 ))(asin(bx)−bcos(bx)) +C
$$\int\:\:\mathrm{e}^{\mathrm{ax}} \:\mathrm{sin}\left(\mathrm{bx}\right)\mathrm{dx}\:\:=\mathrm{Im}\left(\int\:\mathrm{e}^{\mathrm{ax}+\mathrm{ibx}} \mathrm{dx}\right)\:\mathrm{and} \\ $$$$\int\:\mathrm{e}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} \mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{a}+\mathrm{ib}}\:\mathrm{e}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} \:+\mathrm{c} \\ $$$$=\frac{\mathrm{a}−\mathrm{ib}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\:.\mathrm{e}^{\mathrm{ax}} \left\{\mathrm{cos}\left(\mathrm{bx}\right)+\mathrm{isin}\left(\mathrm{bx}\right)\right\}\:\:+\mathrm{c} \\ $$$$=\frac{\mathrm{e}^{\mathrm{ax}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\left\{\mathrm{acos}\left(\mathrm{bx}\right)+\mathrm{iasin}\left(\mathrm{bx}\right)−\mathrm{ibcos}\left(\mathrm{bx}\right)\:+\mathrm{bsin}\left(\mathrm{bx}\right)\right\}\:\Rightarrow \\ $$$$\int\:\mathrm{e}^{\mathrm{ax}} \:\mathrm{sin}\left(\mathrm{bx}\right)\mathrm{dx}\:=\frac{\mathrm{e}^{\mathrm{ax}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\left(\mathrm{asin}\left(\mathrm{bx}\right)−\mathrm{bcos}\left(\mathrm{bx}\right)\right)\:+\mathrm{C} \\ $$

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