Question Number 66334 by mathmax by abdo last updated on 12/Aug/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{e}^{−\mathrm{2}{t}} \:{cos}^{\mathrm{4}} {t}\:{dt}\:{and}\:{J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{−\mathrm{2}{t}} \:{sin}^{\mathrm{4}} {tdt} \\ $$$$\left.\mathrm{1}\right){calculate}\:\:{I}+{J}\:{and}\:{I}−{J} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{I}\:{and}\:{J}. \\ $$
Commented by mathmax by abdo last updated on 13/Aug/19
![1) I+J =∫_0 ^(π/4) (cos^4 t +sin^4 t)e^(−2t) dt =∫_0 ^(π/4) ((cos^2 t+sin^2 t)^2 −2sin^2 tcos^2 t)e^(−2t) dt =∫_0 ^(π/4) (1−(1/2)sin^2 (2t))e^(−2t) dt =∫_0 ^(π/4) e^(−2t) dt−(1/4)∫_0 ^(π/4) (1−cos(4t))e^(−2t) dt =(3/4) ∫_0 ^(π/4) e^(−2t) dt +(1/4) ∫_0 ^(π/4) cos(4t)e^(−2t) dt ∫_0 ^(π/4) e^(−2t) dt =[−(1/2)e^(−2t) ]_0 ^(π/4) =−(1/2)(e^(−(π/2)) −1) ∫_0 ^(π/4) cos(4t)e^(−2t) dt =Re(∫_0 ^(π/4) e^(−2t+i4t) dt) ∫_0 ^(π/4) e^((−2+4i)t) dt =[(1/(−2+4i))e^((−2+4i)t) ]_0 ^(π/4) =−(1/(2−4i)) { e^((−2+4i)(π/4)) −1} =−(((2+4i))/(20)){ e^(−(π/2)) (cos(π)+isin(π))−1} =((1+2i)/(10)){ 1+e^(−(π/2)) } ⇒∫_0 ^(π/4) cos(4t)e^(−2t) dt =(1/(10))e^(−(π/2)) ⇒ I +J =−(3/8)(e^(−(π/2)) −1) +(1/(40)) e^(−(π/2)) =((1/(40))−(3/8))e^(−(π/2)) +(3/8) =−((14)/(40))e^(−(π/2)) +(3/8) =−(7/(20))e^(−(π/2)) +(3/8)](Q66389.png)
$$\left.\mathrm{1}\right)\:{I}+{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left({cos}^{\mathrm{4}} {t}\:+{sin}^{\mathrm{4}} {t}\right){e}^{−\mathrm{2}{t}} {dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\left({cos}^{\mathrm{2}} {t}+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {tcos}^{\mathrm{2}} {t}\right){e}^{−\mathrm{2}{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right){e}^{−\mathrm{2}{t}} {dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{−\mathrm{2}{t}} {dt}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}−{cos}\left(\mathrm{4}{t}\right)\right){e}^{−\mathrm{2}{t}} {dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{−\mathrm{2}{t}} \:{dt}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{4}{t}\right){e}^{−\mathrm{2}{t}} {dt}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{−\mathrm{2}{t}} {dt}\:\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=−\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{−\frac{\pi}{\mathrm{2}}} −\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{4}{t}\right){e}^{−\mathrm{2}{t}} {dt}\:={Re}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{−\mathrm{2}{t}+{i}\mathrm{4}{t}} {dt}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\left(−\mathrm{2}+\mathrm{4}{i}\right){t}} {dt}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{4}{i}}{e}^{\left(−\mathrm{2}+\mathrm{4}{i}\right){t}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=−\frac{\mathrm{1}}{\mathrm{2}−\mathrm{4}{i}}\:\left\{\:{e}^{\left(−\mathrm{2}+\mathrm{4}{i}\right)\frac{\pi}{\mathrm{4}}} −\mathrm{1}\right\} \\ $$$$=−\frac{\left(\mathrm{2}+\mathrm{4}{i}\right)}{\mathrm{20}}\left\{\:{e}^{−\frac{\pi}{\mathrm{2}}} \left({cos}\left(\pi\right)+{isin}\left(\pi\right)\right)−\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}+\mathrm{2}{i}}{\mathrm{10}}\left\{\:\mathrm{1}+{e}^{−\frac{\pi}{\mathrm{2}}} \right\}\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{4}{t}\right){e}^{−\mathrm{2}{t}} {dt}\:=\frac{\mathrm{1}}{\mathrm{10}}{e}^{−\frac{\pi}{\mathrm{2}}} \:\Rightarrow \\ $$$${I}\:+{J}\:=−\frac{\mathrm{3}}{\mathrm{8}}\left({e}^{−\frac{\pi}{\mathrm{2}}} −\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{40}}\:{e}^{−\frac{\pi}{\mathrm{2}}} \:=\left(\frac{\mathrm{1}}{\mathrm{40}}−\frac{\mathrm{3}}{\mathrm{8}}\right){e}^{−\frac{\pi}{\mathrm{2}}} \:+\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$=−\frac{\mathrm{14}}{\mathrm{40}}{e}^{−\frac{\pi}{\mathrm{2}}} \:+\frac{\mathrm{3}}{\mathrm{8}}\:=−\frac{\mathrm{7}}{\mathrm{20}}{e}^{−\frac{\pi}{\mathrm{2}}} \:+\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by mathmax by abdo last updated on 13/Aug/19
![we have I−J =∫_0 ^(π/4) (cos^4 t−sin^4 t)e^(−2t) dt =∫_0 ^(π/4) (cos^2 t−sin^2 t)e^(−2t) dt =∫_0 ^(π/4) cos(2t)e^(−2t) dt =Re(∫_0 ^(π/4) e^(−2t+i2t) dt) and ∫_0 ^(π/4) e^((−2+2i)t) dt =[(1/(−2+2i))e^((−2+2i)t) ]_0 ^(π/4) =−(1/(2−2i)){ e^((−2+2i)(π/4)) −1} =−((2+2i)/8){ e^(−(π/2)) i −1} =((1+i)/4)(1−i e^(−(π/2)) ) =((1−ie^(−(π/2)) +i+e^(−(π/2)) )/4) =((1+e^(−(π/2)) +i(1−e^(−(π/2)) ))/4) ⇒ ∫_0 ^(π/4) cos(2t)e^(−2t) dt =(1/4)(1+e^(−(π/2)) ) ⇒ I−J =(1/4) +(1/4)e^(−(π/2)) we have I+J =(3/8)−(7/(20))e^(−(π/2)) ⇒2I =(1/4)+(3/8) +((1/4)−(7/(20)))e^(−(π/2)) =(5/8) −(1/(10))e^(−(π/2)) also 2J =(3/8)−(1/4) +(−(7/(20))−(1/4))e^(−(π/2)) =(1/8) −(3/5)e^(−(π/2)) ⇒ I =(5/(16)) −(1/(20))e^(−(π/2)) and J =(1/(16)) −(3/(10))e^(−(π/2)) .](Q66392.png)
$${we}\:{have}\:{I}−{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cos}^{\mathrm{4}} {t}−{sin}^{\mathrm{4}} {t}\right){e}^{−\mathrm{2}{t}} {dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cos}^{\mathrm{2}} {t}−{sin}^{\mathrm{2}} {t}\right){e}^{−\mathrm{2}{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{t}\right){e}^{−\mathrm{2}{t}} {dt}\:={Re}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{−\mathrm{2}{t}+{i}\mathrm{2}{t}} \:{dt}\right)\:{and} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right){t}} {dt}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{2}{i}}{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right){t}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} =−\frac{\mathrm{1}}{\mathrm{2}−\mathrm{2}{i}}\left\{\:{e}^{\left(−\mathrm{2}+\mathrm{2}{i}\right)\frac{\pi}{\mathrm{4}}} −\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{2}+\mathrm{2}{i}}{\mathrm{8}}\left\{\:{e}^{−\frac{\pi}{\mathrm{2}}} {i}\:−\mathrm{1}\right\}\:=\frac{\mathrm{1}+{i}}{\mathrm{4}}\left(\mathrm{1}−{i}\:{e}^{−\frac{\pi}{\mathrm{2}}} \right) \\ $$$$=\frac{\mathrm{1}−{ie}^{−\frac{\pi}{\mathrm{2}}} +{i}+{e}^{−\frac{\pi}{\mathrm{2}}} }{\mathrm{4}}\:=\frac{\mathrm{1}+{e}^{−\frac{\pi}{\mathrm{2}}} \:+{i}\left(\mathrm{1}−{e}^{−\frac{\pi}{\mathrm{2}}} \right)}{\mathrm{4}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}{t}\right){e}^{−\mathrm{2}{t}} {dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{e}^{−\frac{\pi}{\mathrm{2}}} \right)\:\Rightarrow\:{I}−{J}\:=\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\frac{\pi}{\mathrm{2}}} \\ $$$${we}\:{have}\:{I}+{J}\:=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{7}}{\mathrm{20}}{e}^{−\frac{\pi}{\mathrm{2}}} \:\Rightarrow\mathrm{2}{I}\:=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{8}}\:+\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{7}}{\mathrm{20}}\right){e}^{−\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{10}}{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\:{also}\:\mathrm{2}{J}\:=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\:+\left(−\frac{\mathrm{7}}{\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{4}}\right){e}^{−\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:−\frac{\mathrm{3}}{\mathrm{5}}{e}^{−\frac{\pi}{\mathrm{2}}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{5}}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{20}}{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\:\:{and}\:{J}\:=\frac{\mathrm{1}}{\mathrm{16}}\:−\frac{\mathrm{3}}{\mathrm{10}}{e}^{−\frac{\pi}{\mathrm{2}}} \:. \\ $$$$ \\ $$