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Question Number 83961 by mathmax by abdo last updated on 08/Mar/20

find I =∫ e^(−x) cos^4 xdx and J =∫ e^(−x) sin^4 xdx

$${find}\:{I}\:=\int\:{e}^{−{x}} \:{cos}^{\mathrm{4}} {xdx}\:\:{and}\:{J}\:=\int\:{e}^{−{x}} \:{sin}^{\mathrm{4}} \:{xdx} \\ $$

Commented by mathmax by abdo last updated on 11/Mar/20

we have I+J =∫ e^(−x) (cos^4 x+sin^4 x)dx =∫ e^(−x) ( (cos^2 x+sin^2 x)^2 −2cos^2 x sin^2 x)dx =∫ e^(−x) (1−2(((sin(2x))/2))^2 )dx =∫ e^(−x) (1−(1/2)sin^2 (2x))dx =∫ e^(−x) dx−(1/2) ∫ e^(−x) ((1−cos(4x))/2)dx =−e^(−x) −(1/4) ∫ e^(−x) dx +(1/4) ∫ e^(−x) cos(4x)dx =(−1+(1/4))e^(−x) +(1/4)Re(∫ e^(−x+i4x) dx) ∫ e^((−1+4i)x) dx =(1/(−1+4i))e^((−1+4i)x) =−(1/(1−4i))e^((−1+4i)x) =−((1+4i)/(17))e^(−x) (cos(4x)+isin(4x)) =−(e^(−x) /(17))(cos(4x)+isin(4x)+4icos(4x)−4sin(4x)) ⇒ I+J =−(3/4)e^(−x) −(e^(−x) /(68))(cos(4x)−4sin(4x)) I−J =∫ e^(−x) (cos^4 x−sin^2 x)dx =∫ e^(−x) (cos^2 x−sin^2 x)dx =∫ e^(−x) cos(2x)dx =Re(∫ e^((−1+2i)x) dx) we have ∫ e^((−1+2i)x) dx =(1/(−1+2i))e^((−1+2i)x) =−((1+2i)/5)e^(−x) ( cos(2x)+isin(2x)) =−(e^(−x) /5)( cos(2x)+isin(2x)+2icos(2x)−2sin(2x)) ⇒ I−J =−(e^(−x) /5)(cos(2x)−2sin(2x)) nowits eazy to determine I and J

$${we}\:{have}\:{I}+{J}\:=\int\:{e}^{−{x}} \left({cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}\right){dx} \\ $$$$=\int\:{e}^{−{x}} \left(\:\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}\right){dx} \\ $$$$=\int\:{e}^{−{x}} \left(\mathrm{1}−\mathrm{2}\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \right){dx}\:=\int\:{e}^{−{x}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\int\:{e}^{−{x}} \:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{e}^{−{x}} \frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=−{e}^{−{x}} −\frac{\mathrm{1}}{\mathrm{4}}\:\int\:{e}^{−{x}} \:{dx}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\:{e}^{−{x}} \:{cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right){e}^{−{x}} \:+\frac{\mathrm{1}}{\mathrm{4}}{Re}\left(\int\:{e}^{−{x}+{i}\mathrm{4}{x}} {dx}\right) \\ $$$$\int\:{e}^{\left(−\mathrm{1}+\mathrm{4}{i}\right){x}} {dx}\:=\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{4}{i}}{e}^{\left(−\mathrm{1}+\mathrm{4}{i}\right){x}} \:\:=−\frac{\mathrm{1}}{\mathrm{1}−\mathrm{4}{i}}{e}^{\left(−\mathrm{1}+\mathrm{4}{i}\right){x}} \\ $$$$=−\frac{\mathrm{1}+\mathrm{4}{i}}{\mathrm{17}}{e}^{−{x}} \left({cos}\left(\mathrm{4}{x}\right)+{isin}\left(\mathrm{4}{x}\right)\right) \\ $$$$=−\frac{{e}^{−{x}} }{\mathrm{17}}\left({cos}\left(\mathrm{4}{x}\right)+{isin}\left(\mathrm{4}{x}\right)+\mathrm{4}{icos}\left(\mathrm{4}{x}\right)−\mathrm{4}{sin}\left(\mathrm{4}{x}\right)\right)\:\Rightarrow \\ $$$${I}+{J}\:=−\frac{\mathrm{3}}{\mathrm{4}}{e}^{−{x}} \:−\frac{{e}^{−{x}} }{\mathrm{68}}\left({cos}\left(\mathrm{4}{x}\right)−\mathrm{4}{sin}\left(\mathrm{4}{x}\right)\right) \\ $$$${I}−{J}\:=\int\:{e}^{−{x}} \left({cos}^{\mathrm{4}} {x}−{sin}^{\mathrm{2}} {x}\right){dx}\:=\int\:{e}^{−{x}} \left({cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}\right){dx} \\ $$$$=\int\:{e}^{−{x}} \:{cos}\left(\mathrm{2}{x}\right){dx}\:={Re}\left(\int\:{e}^{\left(−\mathrm{1}+\mathrm{2}{i}\right){x}} {dx}\right) \\ $$$${we}\:{have}\:\int\:{e}^{\left(−\mathrm{1}+\mathrm{2}{i}\right){x}} {dx}\:=\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{2}{i}}{e}^{\left(−\mathrm{1}+\mathrm{2}{i}\right){x}} \\ $$$$=−\frac{\mathrm{1}+\mathrm{2}{i}}{\mathrm{5}}{e}^{−{x}} \left(\:{cos}\left(\mathrm{2}{x}\right)+{isin}\left(\mathrm{2}{x}\right)\right) \\ $$$$=−\frac{{e}^{−{x}} }{\mathrm{5}}\left(\:\:{cos}\left(\mathrm{2}{x}\right)+{isin}\left(\mathrm{2}{x}\right)+\mathrm{2}{icos}\left(\mathrm{2}{x}\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)\right)\:\Rightarrow \\ $$$${I}−{J}\:=−\frac{{e}^{−{x}} }{\mathrm{5}}\left({cos}\left(\mathrm{2}{x}\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)\right) \\ $$$${nowits}\:{eazy}\:{to}\:{determine}\:{I}\:{and}\:{J} \\ $$

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