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Question Number 83824 by Power last updated on 06/Mar/20
Commented by mathmax by abdo last updated on 06/Mar/20
changement x=(π/2)−t give Ω =−∫_0 ^(π/2) ((π((π/2)−t)+cos^4 t−sin^4 t)/(cos^4 t+sin^4 t))(−dt) =(π^2 /2)∫_0 ^(π/2) (dt/(cos^4 t+sin^4 tdt)) +∫_0 ^(π/2) ((−πt +cos^4 t−sin^4 t)/(cos^4 t +sin^4 t))dt =(π^2 /2)∫_0 ^(π/2) (dt/(cos^4 t +sin^4 t)) −Ω ⇒2Ω =(π^2 /2) ∫_0 ^(π/2) (dt/(1−2cos^2 t sin^2 t)) ⇒ Ω =(π^2 /4)∫_0 ^(π/2) (dt/(1−2(((sin(2t))/2))^2 )) =(π^2 /4)∫_0 ^(π/2) (dt/(1−(1/2)sin^2 (2t))) =(π^2 /2)∫_0 ^(π/2) (dt/(2−((1−cos(4t))/2))) =π^2 ∫_0 ^(π/2) (dt/(3+cos(4t))) =_(4t =u) π^2 ∫_0 ^(2π) (1/(3+cosu))(du/4) =(π^2 /4) ∫_0 ^(2π) (du/(3+cosu)) changement e^(iu) =z ⇒ ∫_0 ^(2π) (du/(3+cosu)) =∫_(∣z∣=1) (1/(3+((z+z^(−1) )/2)))(dz/(iz)) =∫_(∣z∣=1) ((2dz)/(iz(6+z+z^(−1) ))) =∫_(∣z∣=1) ((−2idz)/(6z +z^2 +1)) let W(z)=((−2i)/(z^2 +6z +1)) poles of W? Δ^′ =9−1 =8 ⇒z_1 =−3+2(√2) and z_2 =−3−2(√2) ⇒W(z)=((−2i)/((z−z_1 )(z−z_2 ))) ∣z_1 ∣−1 =∣−3+2(√2)∣−1 =3−2(√2)−1 =2−2(√2)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣ =3+2(√2)−1 =2+2(√2)>1 residus theorem give ∫_(∣z∣=1) W(z)dz =2iπ Res(W,z_1 ) =2iπ×((−2i)/(z_1 −z_2 )) =((4π)/(4(√2))) =(π/( (√2))) ⇒ Ω =(π^2 /4)×(π/( (√2))) =(π^3 /(4(√2)))
$${changement}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\:{give} \\ $$$$\Omega\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\pi\left(\frac{\pi}{\mathrm{2}}−{t}\right)+{cos}^{\mathrm{4}} {t}−{sin}^{\mathrm{4}} {t}}{{cos}^{\mathrm{4}} {t}+{sin}^{\mathrm{4}} {t}}\left(−{dt}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{cos}^{\mathrm{4}} {t}+{sin}^{\mathrm{4}} \:{tdt}}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{−\pi{t}\:+{cos}^{\mathrm{4}} {t}−{sin}^{\mathrm{4}} {t}}{{cos}^{\mathrm{4}} {t}\:+{sin}^{\mathrm{4}} {t}}{dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{{cos}^{\mathrm{4}} {t}\:+{sin}^{\mathrm{4}} {t}}\:−\Omega\:\Rightarrow\mathrm{2}\Omega\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} {t}\:{sin}^{\mathrm{2}} {t}}\:\Rightarrow \\ $$$$\Omega\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{1}−\mathrm{2}\left(\frac{{sin}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right)} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{2}−\frac{\mathrm{1}−{cos}\left(\mathrm{4}{t}\right)}{\mathrm{2}}}\:=\pi^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{3}+{cos}\left(\mathrm{4}{t}\right)} \\ $$$$=_{\mathrm{4}{t}\:={u}} \:\:\pi^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{\mathrm{3}+{cosu}}\frac{{du}}{\mathrm{4}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{du}}{\mathrm{3}+{cosu}}\:\:{changement}\:{e}^{{iu}} ={z}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{du}}{\mathrm{3}+{cosu}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{3}+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\frac{{dz}}{{iz}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{6}+{z}+{z}^{−\mathrm{1}} \right)}\:=\int_{\mid{z}\mid=\mathrm{1}} \frac{−\mathrm{2}{idz}}{\mathrm{6}{z}\:+{z}^{\mathrm{2}} +\mathrm{1}}\:\:{let}\:{W}\left({z}\right)=\frac{−\mathrm{2}{i}}{{z}^{\mathrm{2}} +\mathrm{6}{z}\:+\mathrm{1}} \\ $$$${poles}\:{of}\:{W}? \\ $$$$\Delta^{'} =\mathrm{9}−\mathrm{1}\:=\mathrm{8}\:\Rightarrow{z}_{\mathrm{1}} =−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:{and}\:{z}_{\mathrm{2}} =−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:\Rightarrow{W}\left({z}\right)=\frac{−\mathrm{2}{i}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\mid−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\mid−\mathrm{1}\:=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\:=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\:=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}>\mathrm{1}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{z}_{\mathrm{1}} \right)\:=\mathrm{2}{i}\pi×\frac{−\mathrm{2}{i}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\frac{\mathrm{4}\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\Omega\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}×\frac{\pi}{\:\sqrt{\mathrm{2}}}\:=\frac{\pi^{\mathrm{3}} }{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$
Answered by TANMAY PANACEA last updated on 06/Mar/20
I=∫_0 ^(π/2) ((π((π/2)−x)+sin^4 ((π/2)−x)−cos^4 ((π/2)−x))/(sin^4 ((π/2)−x)+cos^4 ((π/2)−x)))dx 2I=∫_0 ^(π/2) ((π((π/2)))/(cos^4 x+sin^4 x)) 2I=(π^2 /2)∫_0 ^(π/2) ((sec^2 x×(1+tan^2 x))/(1+tan^4 x))dx I=(π^2 /4)∫_0 ^∞ ((1+t^2 )/(1+t^4 ))dt [t=tanx] I=(π^2 /4)∫_0 ^∞ ((1+(1/t^2 ))/(t^2 +(1/t^2 )))dt =(π^2 /4)∫_0 ^∞ ((d(t−(1/t)))/((t−(1/t))^2 +2)) (π^2 /4)×(1/( (√2)))∣tan^(−1) (((t−(1/t))/( (√2))))∣_0 ^∞ (π^2 /(4(√2)))×{(tan^(−1) ∞−tan^(−1) (−∞)} (π^2 /(4(√2)))×((π/2)+(π/2))=(π^3 /(4(√2))) pls check
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\pi\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)−{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)+{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\pi\left(\frac{\pi}{\mathrm{2}}\right)}{{cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}} \\ $$$$\mathrm{2}{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {x}×\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)}{\mathrm{1}+{tan}^{\mathrm{4}} {x}}{dx} \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:\left[{t}={tanx}\right] \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{2}}}×\left\{\left({tan}^{−\mathrm{1}} \infty−{tan}^{−\mathrm{1}} \left(−\infty\right)\right\}\right. \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{2}}}×\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{3}} }{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$
Commented by mathmax by abdo last updated on 06/Mar/20
your answer is correct sir tanmay
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{tanmay} \\ $$
Commented by Power last updated on 06/Mar/20
thanks
$$\mathrm{thanks} \\ $$
Commented by TANMAY PANACEA last updated on 06/Mar/20
thank you both of you sir
$${thank}\:{you}\:{both}\:{of}\:{you}\:{sir} \\ $$
Answered by Kamel Kamel last updated on 07/Mar/20
Put t=(π/2)−x ∴ Ω=∫_0 ^(π/2) (((π^2 /2)−πt+cos^4 (t)−sin^4 (t))/(cos^4 (t)+sin^4 (t)))dx ∴ 2Ω=(π^2 /2)∫_0 ^(π/2) (dx/(cos^4 (x)+sin^4 (x)))=(π^2 /2)∫_0 ^(π/2) (dx/(1−2sin^2 (x)cos^2 (x))) =2π^2 ∫_0 ^(π/2) (dx/(2−sin^2 (2x)))=^(t=2x) π^2 ∫_0 ^π (dt/(2−sin^2 (t)))=2π^2 ∫_0 ^π (dt/(cos(2t)+3)) =^(u=2t) π^2 ∫_0 ^(2π) (du/(cos(u)+3))=π^2 ∫_(−π) ^π (du/(cos(u)+3))=^(v=tg((u/2)) 2π^2 ∫_(−∞) ^(+∞) (dv/(1−v^2 +3+3v^2 )) =2π^2 ∫_0 ^(+∞) (dv/(v^2 +2))=(π^2 /( (√2)))Arctg((v/( (√2))))∣_0 ^(+∞) =(π^3 /(2(√2)))⇒Ω=(π^3 /(4(√2))) ∴ ∫_0 ^(π/2) ((πx+sin^4 (x)−cos^4 (x))/(sin^4 (x)+cos^4 (x)))dx=(π^3 /(4(√2))) KAMEL BENAICHA
$${Put}\:{t}=\frac{\pi}{\mathrm{2}}−{x} \\ $$$$\therefore\:\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi{t}+{cos}^{\mathrm{4}} \left({t}\right)−{sin}^{\mathrm{4}} \left({t}\right)}{{cos}^{\mathrm{4}} \left({t}\right)+{sin}^{\mathrm{4}} \left({t}\right)}{dx} \\ $$$$\therefore\:\mathrm{2}\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{cos}^{\mathrm{4}} \left({x}\right)+{sin}^{\mathrm{4}} \left({x}\right)}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{2}−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}\overset{{t}=\mathrm{2}{x}} {=}\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\pi} \frac{{dt}}{\mathrm{2}−{sin}^{\mathrm{2}} \left({t}\right)}=\mathrm{2}\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\pi} \frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)+\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\overset{{u}=\mathrm{2}{t}} {=}\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{du}}{{cos}\left({u}\right)+\mathrm{3}}=\pi^{\mathrm{2}} \int_{−\pi} ^{\pi} \frac{{du}}{{cos}\left({u}\right)+\mathrm{3}}\overset{{v}={tg}\left(\frac{{u}}{\mathrm{2}}\right.} {=}\:\:\:\:\mathrm{2}\pi^{\mathrm{2}} \int_{−\infty} ^{+\infty} \frac{{dv}}{\mathrm{1}−{v}^{\mathrm{2}} +\mathrm{3}+\mathrm{3}{v}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{+\infty} \frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}}{Arctg}\left(\frac{{v}}{\:\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{+\infty} =\frac{\pi^{\mathrm{3}} }{\mathrm{2}\sqrt{\mathrm{2}}}\Rightarrow\Omega=\frac{\pi^{\mathrm{3}} }{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\pi{x}+{sin}^{\mathrm{4}} \left({x}\right)−{cos}^{\mathrm{4}} \left({x}\right)}{{sin}^{\mathrm{4}} \left({x}\right)+{cos}^{\mathrm{4}} \left({x}\right)}{dx}=\frac{\pi^{\mathrm{3}} }{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\ $$

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