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Question Number 64649 by mathmax by abdo last updated on 20/Jul/19
calculate ∫_0 ^(2π)   ((cosθ)/(5+3cosθ))dθ
calculate02πcosθ5+3cosθdθ
Commented by mathmax by abdo last updated on 20/Jul/19
let A =∫_0 ^(2π)    ((cosθ)/(5+3cosθ)) ⇒A =(1/3) ∫_0 ^(2π)   ((3cosθ +5−5)/(3cosθ +5))dθ  =((2π)/3) −(5/3) ∫_0 ^(2π)   (dθ/(3cosθ +5))  changement e^(iθ) =z give  ∫_0 ^(2π)   (dθ/(3cosθ +5)) =∫_(∣z∣=1)      (1/(3 ((z+z^(−1) )/2)+5)) (dz/(iz))  = ∫_(∣z∣=1)   ((2dz)/(iz(3z+3z^(−1)  +10))) =∫_(∣z∣=1)    ((−2i)/(3z^2  +3 +10z))dz let  W(z) =((−2i)/(3z^2  +10z +3))  poles of W?  Δ^′ =5^2 −9 =16 ⇒z_1 =((−5 +4)/3) =−(1/3)  and z_2 =((−5−4)/3) =−3 ⇒  W(z) =((−2i)/(3(z−z_1 )(z−z_2 )))     (z_2 is out of circle) residus theorem give  ∫_(∣z∣=1) W(z)dz =2iπ Res(W,z_1 )  Res(W,z_1 ) =lim_(z→z_1 ) (z−z_1 )W(z) =((−2i)/(3(z_1 −z_2 ))) =((−2i)/(3(−(1/3)+3)))  =((−2i)/(3.(8/3))) =−(i/4) ⇒∫_(∣z∣=1) W(z)dz =2iπ(−(i/4)) =(π/2) ⇒  A =((2π)/3) −(5/3) (π/2) =((2π)/3)−((5π)/6) =((4π−5π)/6) ⇒ A =−(π/6) .
letA=02πcosθ5+3cosθA=1302π3cosθ+553cosθ+5dθ=2π35302πdθ3cosθ+5changementeiθ=zgive02πdθ3cosθ+5=z∣=113z+z12+5dziz=z∣=12dziz(3z+3z1+10)=z∣=12i3z2+3+10zdzletW(z)=2i3z2+10z+3polesofW?Δ=529=16z1=5+43=13andz2=543=3W(z)=2i3(zz1)(zz2)(z2isoutofcircle)residustheoremgivez∣=1W(z)dz=2iπRes(W,z1)Res(W,z1)=limzz1(zz1)W(z)=2i3(z1z2)=2i3(13+3)=2i3.83=i4z∣=1W(z)dz=2iπ(i4)=π2A=2π353π2=2π35π6=4π5π6A=π6.
Answered by MJS last updated on 20/Jul/19
∫_0 ^(2π) ((cos θ)/(5+3cos θ))dθ=2∫_0 ^π ((cos θ)/(5+3cos θ))dθ=       [t=tan (θ/2) → dθ=2cos^2  (θ/2) dt]  −2∫_0 ^∞ ((t^2 −1)/((t^2 +1)(t^2 +4)))dt=(4/3)∫_0 ^∞ (dt/(t^2 +1))−((10)/3)∫_0 ^∞ (dt/(t^2 +4))=  =[(4/3)arctan t −(5/3)arctan (t/2)]_0 ^∞ =−(π/6)
2π0cosθ5+3cosθdθ=2π0cosθ5+3cosθdθ=[t=tanθ2dθ=2cos2θ2dt]20t21(t2+1)(t2+4)dt=430dtt2+11030dtt2+4==[43arctant53arctant2]0=π6
Commented by mathmax by abdo last updated on 20/Jul/19
thank you sir.
thankyousir.

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