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let-I-0-pi-8-e-2t-cos-4-t-and-J-0-pi-8-e-2t-sin-4-dt-find-the-values-of-I-andJ-

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Question Number 42799 by maxmathsup by imad last updated on 02/Sep/18
let I = ∫_0 ^(π/8) e^(−2t) cos^4 t and J =∫_0 ^(π/8) e^(−2t) sin^4 dt find the values of I andJ .
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{−\mathrm{2}{t}} \:{cos}^{\mathrm{4}} {t}\:\:\:\:{and}\:{J}\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \:{sin}^{\mathrm{4}} {dt} \\ $$$${find}\:{the}\:{values}\:{of}\:{I}\:{andJ}\:. \\ $$
Commented by maxmathsup by imad last updated on 06/Sep/18
we have I +J = ∫_0 ^(π/8) e^(−2t) { cos^4 t +sin^4 t}dt = ∫_0 ^(π/8) e^(−2t) { (cos^2 t +sin^2 t)^2 −2cos^2 t sin^2 t}dt =∫_0 ^(π/8) e^(−2t) { 1−(1/2) sin^2 2t} dt = ∫_0 ^(π/8) e^(−2t) dt −(1/2) ∫_0 ^(π/8) e^(−2t) sin^2 (2t) dt =−(1/2)[ e^(−2t) ]_0 ^(π/8) −(1/4) ∫_0 ^(π/8) e^(−2t) (1−cos(2t))dt =(1/2){1−e^(−(π/4)) } +(1/8)[ e^(−2t) ]_0 ^(π/8) +(1/4) ∫_0 ^(π/8) e^(−2t) cos(2t)dt =(1/2){1−e^(−(π/4)) } +(1/8){ e^(−(π/4)) −1} +(1/4) Re( ∫_0 ^(π/8) e^(−2t +2it) dt) but ∫_0 ^(π/8) e^((−2+2i)t) dt =[ (1/(−2+2i)) e^((−2+2i)t) ]_0 ^(π/8) =(1/(−2 +2i)) { e^((−2+2i)(π/8)) −1} = −(1/2) (1/(1−i)){ e^(−(π/4)) e^(i(π/4)) −1}} =−(1/2) (((1+i))/2)){ e^(−(π/4)) ( (1/( (√2))) +(i/( (√2)))) −1} =((−1)/4)(1+i) { (e^(−(π/4)) /( (√2))) + i (e^(−(π/4)) /( (√2))) −1} =−(1/(4(√2))){ e^(−(π/4)) +i e^(−(π/4)) −(√2)}(1+i) =−(1/(4(√2))){ e^(−(π/4)) +i e^(−(π/4)) +i e^(−(π/4)) −e^(−(π/4)) −(√2)−(√2)i} = −(1/(4(√2))){ 2i e^(−(π/4)) −(√2)i −(√2)} ⇒Re( ∫_0 ^(π/2) ...dx)= (1/4) ⇒ I +J =(1/2) −(1/8) −(3/8) e^(−(π/4)) +(1/(16)) = (7/(16)) −(3/8) e^(−(π/4)) .... be continued...
$${we}\:{have}\:\:{I}\:+{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \left\{\:{cos}^{\mathrm{4}} {t}\:+{sin}^{\mathrm{4}} {t}\right\}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \left\{\:\left({cos}^{\mathrm{2}} {t}\:+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} \:−\mathrm{2}{cos}^{\mathrm{2}} {t}\:{sin}^{\mathrm{2}} {t}\right\}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{−\mathrm{2}{t}} \left\{\:\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:{sin}^{\mathrm{2}} \mathrm{2}{t}\right\}\:{dt}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} {dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \:\:{sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\:{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:\:−\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \:\left(\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−{e}^{−\frac{\pi}{\mathrm{4}}} \right\}\:\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\:{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \:{cos}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−{e}^{−\frac{\pi}{\mathrm{4}}} \right\}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\:{e}^{−\frac{\pi}{\mathrm{4}}} \:−\mathrm{1}\right\}\:+\frac{\mathrm{1}}{\mathrm{4}}\:{Re}\left(\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{−\mathrm{2}{t}\:+\mathrm{2}{it}} {dt}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right){t}} {dt}\:=\left[\:\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{2}{i}}\:{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right){t}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:=\frac{\mathrm{1}}{−\mathrm{2}\:+\mathrm{2}{i}}\:\left\{\:{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right)\frac{\pi}{\mathrm{8}}} \:−\mathrm{1}\right\} \\ $$$$\left.=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\frac{\mathrm{1}}{\mathrm{1}−{i}}\left\{\:{e}^{−\frac{\pi}{\mathrm{4}}} \:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:−\mathrm{1}\right\}\right\}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\left(\frac{\left.\mathrm{1}+{i}\right)}{\mathrm{2}}\right)\left\{\:{e}^{−\frac{\pi}{\mathrm{4}}} \left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:−\mathrm{1}\right\} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{i}\right)\:\left\{\:\frac{{e}^{−\frac{\pi}{\mathrm{4}}} }{\:\sqrt{\mathrm{2}}}\:+\:{i}\:\frac{{e}^{−\frac{\pi}{\mathrm{4}}} }{\:\sqrt{\mathrm{2}}}\:−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left\{\:{e}^{−\frac{\pi}{\mathrm{4}}} \:+{i}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:−\sqrt{\mathrm{2}}\right\}\left(\mathrm{1}+{i}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left\{\:\:{e}^{−\frac{\pi}{\mathrm{4}}} \:+{i}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:\:+{i}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:−{e}^{−\frac{\pi}{\mathrm{4}}} \:−\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}{i}\right\} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left\{\:\mathrm{2}{i}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:−\sqrt{\mathrm{2}}{i}\:−\sqrt{\mathrm{2}}\right\}\:\Rightarrow{Re}\left(\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:…{dx}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\: \\ $$$${I}\:+{J}\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{8}}\:−\frac{\mathrm{3}}{\mathrm{8}}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:\:+\frac{\mathrm{1}}{\mathrm{16}}\:=\:\frac{\mathrm{7}}{\mathrm{16}}\:−\frac{\mathrm{3}}{\mathrm{8}}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:….\:{be}\:{continued}… \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 06/Sep/18
we have I −J = ∫_0 ^(π/8) e^(−2t) {cos^4 t −sin^4 t}dt =∫_0 ^(π/8) e^(−2t) {cos^2 t −sin^2 t}dt = ∫_0 ^(π/8) e^(−2t) cos(2t)dt =Re( ∫_0 ^(π/8) e^(−2t +2it) dt) =Re( ∫_0 ^(π/8) e^((−2+2i)t) dt)=(1/4) so I+J =(7/(16)) −(3/8)e^(−(π/4)) and I−J=(1/4) ⇒ 2I =((11)/(16)) −(3/8) e^(−(π/4)) and 2J =(3/(16)) −(3/8) e^(−(π/4)) ⇒ I =((11)/(32)) −(3/(16)) e^(−(π/4)) and J =(3/(32)) −(3/(16)) e^(−(π/4))
$${we}\:{have}\:{I}\:−{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \left\{{cos}^{\mathrm{4}} {t}\:−{sin}^{\mathrm{4}} {t}\right\}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \left\{{cos}^{\mathrm{2}} {t}\:−{sin}^{\mathrm{2}} {t}\right\}{dt}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{e}^{−\mathrm{2}{t}} \:{cos}\left(\mathrm{2}{t}\right){dt}\:={Re}\left(\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{−\mathrm{2}{t}\:+\mathrm{2}{it}} {dt}\right) \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\:{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right){t}} {dt}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:\:{so}\:\:{I}+{J}\:=\frac{\mathrm{7}}{\mathrm{16}}\:−\frac{\mathrm{3}}{\mathrm{8}}{e}^{−\frac{\pi}{\mathrm{4}}} \:\:{and}\:{I}−{J}=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\frac{\mathrm{11}}{\mathrm{16}}\:−\frac{\mathrm{3}}{\mathrm{8}}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:\:\:{and}\:\mathrm{2}{J}\:=\frac{\mathrm{3}}{\mathrm{16}}\:−\frac{\mathrm{3}}{\mathrm{8}}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{11}}{\mathrm{32}}\:−\frac{\mathrm{3}}{\mathrm{16}}\:{e}^{−\frac{\pi}{\mathrm{4}}} \:\:{and}\:{J}\:=\frac{\mathrm{3}}{\mathrm{32}}\:−\frac{\mathrm{3}}{\mathrm{16}}\:{e}^{−\frac{\pi}{\mathrm{4}}} \\ $$

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