cos3x-e-2x-dx-

Question Number 76462 by kaivan.ahmadi last updated on 27/Dec/19

$$\int{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} {dx} \\ $$

Commented by mathmax by abdo last updated on 27/Dec/19

let A =∫ cos(3x)e^(−2x) dx ⇒A =Re(∫ e^(3ix) e^(−2x) dx) we have ∫ e^((−2+3i)x) dx =(1/(−2+3i)) e^((−2+3i)x) +c =((3i+2)/((3i)^2 −2^2 )) e^(−2x) { cos(3x)+isin(3x)} +c =−(1/(13)) e^(−2x) (3i+2){cos(3x)+isin(3x)} +c =−(1/(13))e^(−2x) { 3i cos(3x)−3sin(3x)+2cos(3x)+2i sin(3x)} ⇒ A =−(1/(13))e^(−2x) {2cos(3x)−3sin(3x)} +c =(1/(13))e^(−2x) {3sin(3x)+2cos(3x)} +c

$${let}\:{A}\:=\int\:{cos}\left(\mathrm{3}{x}\right){e}^{−\mathrm{2}{x}} {dx}\:\Rightarrow{A}\:={Re}\left(\int\:{e}^{\mathrm{3}{ix}} \:{e}^{−\mathrm{2}{x}} {dx}\right)\:{we}\:{have} \\ $$$$\int\:{e}^{\left(−\mathrm{2}+\mathrm{3}{i}\right){x}} {dx}\:=\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{3}{i}}\:{e}^{\left(−\mathrm{2}+\mathrm{3}{i}\right){x}} \:+{c} \\ $$$$=\frac{\mathrm{3}{i}+\mathrm{2}}{\left(\mathrm{3}{i}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }\:{e}^{−\mathrm{2}{x}} \left\{\:{cos}\left(\mathrm{3}{x}\right)+{isin}\left(\mathrm{3}{x}\right)\right\}\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{13}}\:{e}^{−\mathrm{2}{x}} \left(\mathrm{3}{i}+\mathrm{2}\right)\left\{{cos}\left(\mathrm{3}{x}\right)+{isin}\left(\mathrm{3}{x}\right)\right\}\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{13}}{e}^{−\mathrm{2}{x}} \left\{\:\mathrm{3}{i}\:{cos}\left(\mathrm{3}{x}\right)−\mathrm{3}{sin}\left(\mathrm{3}{x}\right)+\mathrm{2}{cos}\left(\mathrm{3}{x}\right)+\mathrm{2}{i}\:{sin}\left(\mathrm{3}{x}\right)\right\}\:\Rightarrow \\ $$$${A}\:=−\frac{\mathrm{1}}{\mathrm{13}}{e}^{−\mathrm{2}{x}} \left\{\mathrm{2}{cos}\left(\mathrm{3}{x}\right)−\mathrm{3}{sin}\left(\mathrm{3}{x}\right)\right\}\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{13}}{e}^{−\mathrm{2}{x}} \left\{\mathrm{3}{sin}\left(\mathrm{3}{x}\right)+\mathrm{2}{cos}\left(\mathrm{3}{x}\right)\right\}\:+{c} \\ $$

Commented by john santu last updated on 28/Dec/19

$${this}\:{is}\:{using}\:{De}\:{Moivre}'{s}\:{theorem}\:{sir}? \\ $$

Commented by mathmax by abdo last updated on 28/Dec/19

not moivre but 2Re(z)=z+z^− and 2iIm(z)=z−z^−

$${not}\:{moivre}\:{but}\:\:\mathrm{2}{Re}\left({z}\right)={z}+\overset{−} {{z}}\:\:\:{and}\:\mathrm{2}{iIm}\left({z}\right)={z}−\overset{−} {{z}} \\ $$

Commented by kaivan.ahmadi last updated on 29/Dec/19

$${thank}\:{you}\:{sir} \\ $$

Answered by Kunal12588 last updated on 27/Dec/19

I=∫cos 3x e^(−2x) dx =cos3x (e^(−2x) /(−2))−∫−sin 3x(3)(e^(−2x) /(−2))dx =−(1/2)cos3x e^(−2x) −(3/2)∫sin 3x e^(−2x) dx I_1 =∫sin 3x e^(−2x) =−(1/2)sin 3x e^(−2x) +(3/2)∫cos 3x e^(−2x) dx =−(1/2)sin 3x e^(−2x) +(3/2)I I=−(1/2)cos3x e^(−2x) −(3/2)(−(1/2)sin 3x e^(−2x) +(3/2)I) ⇒I=−(1/2)cos3x e^(−2x) +(3/4)sin 3x e^(−2x) −(9/4)I ⇒((13)/4)I=−(1/2)cos3x e^(−2x) +(3/4)sin 3x e^(−2x) +C ⇒I=(1/(13))(−2cos3x e^(−2x) +3sin 3x e^(−2x) )+C ∫cos 3x e^(−2x) dx=(1/(13))[3sin 3x e^(−2x) −2cos 3x e^(−2x) ]+C

$${I}=\int{cos}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} \:{dx} \\ $$$$={cos}\mathrm{3}{x}\:\frac{{e}^{−\mathrm{2}{x}} }{−\mathrm{2}}−\int−{sin}\:\mathrm{3}{x}\left(\mathrm{3}\right)\frac{{e}^{−\mathrm{2}{x}} }{−\mathrm{2}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} −\frac{\mathrm{3}}{\mathrm{2}}\int{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} {dx} \\ $$$${I}_{\mathrm{1}} =\int{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +\frac{\mathrm{3}}{\mathrm{2}}\int{cos}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} \:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +\frac{\mathrm{3}}{\mathrm{2}}{I} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} −\frac{\mathrm{3}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +\frac{\mathrm{3}}{\mathrm{2}}{I}\right) \\ $$$$\Rightarrow{I}=−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +\frac{\mathrm{3}}{\mathrm{4}}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} −\frac{\mathrm{9}}{\mathrm{4}}{I} \\ $$$$\Rightarrow\frac{\mathrm{13}}{\mathrm{4}}{I}=−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +\frac{\mathrm{3}}{\mathrm{4}}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +{C} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{13}}\left(−\mathrm{2}{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} +\mathrm{3}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} \right)+{C} \\ $$$$\int{cos}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} {dx}=\frac{\mathrm{1}}{\mathrm{13}}\left[\mathrm{3}{sin}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} −\mathrm{2}{cos}\:\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} \right]+{C} \\ $$

Commented by kaivan.ahmadi last updated on 27/Dec/19

$${thank}\:{u}\:{sir} \\ $$

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