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Question Number 28815 by abdo imad last updated on 30/Jan/18

find the value of ∫_0 ^π (dx/(2cos^2 x +sin^2 x)) .

$${find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}}\:. \\ $$

Answered by mrW2 last updated on 31/Jan/18

∫_0 ^π (dx/(2cos^2 x +sin^2 x)) =∫_0 ^π (dx/(1+cos^2 x)) =2∫_0 ^π (dx/(3+2cos^2 x−1)) =2∫_0 ^π (dx/(3+cos 2x)) =2×2[((tan^(−1) (((tan x)/(√2))))/(2(√2)))]_0 ^(π/2) =(π/(√2))

$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}}\: \\ $$$$=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\: \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{3}+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}\: \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{3}+\mathrm{cos}\:\mathrm{2}{x}}\: \\ $$$$=\mathrm{2}×\mathrm{2}\left[\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:{x}}{\sqrt{\mathrm{2}}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{2}}} \\ $$

Commented by abdo imad last updated on 31/Jan/18

another metod by redidus theorem we have I = ∫_0 ^π (dx/(1+cos^2 x))= ∫_0 ^π (dx/(1+((1+cos(2x))/2)))=∫_0 ^π ((2dx)/(3+cos(2x))) the ch. 2x=t give I= ∫_0 ^(2π) (dt/(3+cost)) and the ch. e^(it) =z give I= ∫_(∣z∣=1) (1/(3+ ((z+z^(−1) )/2))) (dz/(iz))= ∫_(∣z∣=1) ((2dz)/(iz( 6+z+z^(−1) ))) = ∫_(∣z∣) ((−2i)/(6z +z^2 +1))= ∫_(∣z∣=1) ((−2i)/(z^2 +6z +1))dz let put f(z)= ((−2i)/(z^2 +6z +1)) , poles of f? z^2 +6z+1=0⇒Δ=36−4= 32 ⇒z_1 =((−6 +4(√2))/2)=−3+2(√2) z_2 = −3−2(√2) we have ∣z_1 ∣−1=2(√2)−3−1=2((√2)−2)<0 and ∣z_2 ∣−1=3+2(√2)−1=2+2(√2) >1 (to eliminate because out of circle) ∫_(∣z∣=1) f(z)dx=2iπ Re(f ,z_1 ) but f(z)= ((−2i)/((z−z_1 )(z−z_2 ))) Res(f,z_1 )=lim_(z→z_1 ) (z−z_1 )f(z)= ((−2i)/(z_1 −z_2 ))= ((−2i)/(4(√2))) ∫_(∣z∣=1) f(z)dx= 2iπ.((−i)/(2(√2)))= (π/(√2)) so I= (π/(√2)) .

$${another}\:{metod}\:{by}\:{redidus}\:{theorem}\:{we}\:{have} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\mathrm{1}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}{dx}}{\mathrm{3}+{cos}\left(\mathrm{2}{x}\right)} \\ $$$${the}\:{ch}.\:\mathrm{2}{x}={t}\:\:\:{give}\:\:{I}=\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{\mathrm{3}+{cost}}\:\:{and}\:{the}\:{ch}.\:{e}^{{it}} ={z}\:{give} \\ $$$${I}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{3}+\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\:\mathrm{6}+{z}+{z}^{−\mathrm{1}} \right)} \\ $$$$=\:\:\int_{\mid{z}\mid} \:\:\:\frac{−\mathrm{2}{i}}{\mathrm{6}{z}\:+{z}^{\mathrm{2}} +\mathrm{1}}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{i}}{{z}^{\mathrm{2}} \:+\mathrm{6}{z}\:+\mathrm{1}}{dz}\:{let}\:{put} \\ $$$${f}\left({z}\right)=\:\frac{−\mathrm{2}{i}}{{z}^{\mathrm{2}} \:+\mathrm{6}{z}\:+\mathrm{1}}\:\:,\:{poles}\:{of}\:{f}? \\ $$$${z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{1}=\mathrm{0}\Rightarrow\Delta=\mathrm{36}−\mathrm{4}=\:\mathrm{32}\:\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{6}\:+\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}=−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\:−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:{we}\:{have}\:\mid{z}_{\mathrm{1}} \mid−\mathrm{1}=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}−\mathrm{1}=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{2}\right)<\mathrm{0} \\ $$$${and}\:\mid{z}_{\mathrm{2}} \mid−\mathrm{1}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\:>\mathrm{1}\:\left({to}\:{eliminate}\:{because}\right. \\ $$$$\left.{out}\:{of}\:{circle}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dx}=\mathrm{2}{i}\pi\:{Re}\left({f}\:,{z}_{\mathrm{1}} \right)\:{but}\:{f}\left({z}\right)=\:\frac{−\mathrm{2}{i}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left({f},{z}_{\mathrm{1}} \right)={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right){f}\left({z}\right)=\:\frac{−\mathrm{2}{i}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }=\:\frac{−\mathrm{2}{i}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dx}=\:\mathrm{2}{i}\pi.\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}}}=\:\frac{\pi}{\sqrt{\mathrm{2}}}\:{so}\:{I}=\:\frac{\pi}{\sqrt{\mathrm{2}}}\:\:. \\ $$

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