Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$\int\:\:\mathrm{e}^{\mathrm{ax}} \:\mathrm{sin}\left(\mathrm{bx}\right)\mathrm{dx}\:\:=\mathrm{Im}\left(\int\:\mathrm{e}^{\mathrm{ax}+\mathrm{ibx}} \mathrm{dx}\right)\:\mathrm{and} \\ $$$$\int\:\mathrm{e}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} \mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{a}+\mathrm{ib}}\:\mathrm{e}^{\left(\mathrm{a}+\mathrm{ib}\right)\mathrm{x}} \:+\mathrm{c} \\ $$$$=\frac{\mathrm{a}−\mathrm{ib}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\:.\mathrm{e}^{\mathrm{ax}} \left\{\mathrm{cos}\left(\mathrm{bx}\right)+\mathrm{isin}\left(\mathrm{bx}\right)\right\}\:\:+\mathrm{c} \\ $$$$=\frac{\mathrm{e}^{\mathrm{ax}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\left\{\mathrm{acos}\left(\mathrm{bx}\right)+\mathrm{iasin}\left(\mathrm{bx}\right)−\mathrm{ibcos}\left(\mathrm{bx}\right)\:+\mathrm{bsin}\left(\mathrm{bx}\right)\right\}\:\Rightarrow \\ $$$$\int\:\mathrm{e}^{\mathrm{ax}} \:\mathrm{sin}\left(\mathrm{bx}\right)\mathrm{dx}\:=\frac{\mathrm{e}^{\mathrm{ax}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\left(\mathrm{asin}\left(\mathrm{bx}\right)−\mathrm{bcos}\left(\mathrm{bx}\right)\right)\:+\mathrm{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com