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calculate-0-pi-cos-2-x-2-3sin-2x-dx-




Question Number 53080 by Abdo msup. last updated on 17/Jan/19
calculate ∫_0 ^π   ((cos^2 x)/(2+3sin(2x)))dx
calculate0πcos2x2+3sin(2x)dx
Commented by maxmathsup by imad last updated on 17/Jan/19
let A =∫_0 ^π   ((cos^2 x)/(2+3sin(2x)))dx ⇒A=∫_0 ^π   ((1+cos(2x))/(2(2+3sin(2x))))dx  =_(2x=t)  ∫_0 ^(2π)   ((1+cos(t))/(2(2+3sin(t)))) (dt/2) =(1/4) ∫_0 ^(2π)    ((1+cos(t))/(2+3sin(t)))dt ⇒  4A =∫_0 ^(2π)    ((1+cost)/(2+3sint)) dt  = Re( ∫_0 ^(2π)   ((1+e^(it) )/(2+3sint))dt) let  changement e^(it) =z give  ∫_0 ^(2π)   ((1+e^(it) )/(2+3sint))dt =∫_(∣z∣=1)   ((1+z)/(2+3((z−z^(−1) )/(2i))))(dz/(iz)) = ∫_(∣z∣=1)    ((2i(1+z))/(iz(4i +3z−3z^(−1) )))dz  =∫_(∣z∣=1)    ((2(1+z))/(4iz +3z^2  −3))dz  let consider the complex function   ϕ(z)=((2(1+z))/(3z^2  +4iz −3))  poles of ϕ?  let find roots of 3z^2  +4iz −3  Δ^′ =(2i)^2 −3(−3) =−4+9 =5 ⇒z_1 =((−2i+(√5))/3)  and z_2 =((−2i−(√5))/3)(are poles of ϕ)  ∣z_1 ∣−1 =(1/3)(√(4+5))=3 >0 ⇒z_1  is out of circle ⇒Res(ϕ,z_1 )=0  ∣z_2 ∣−1 =(1/3)(√(4+5))=3>0 ⇒z_2 is out of circle ⇒Res(ϕ,z_2 )=0   ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,z_1 ) +Res(ϕ,z_2 )}=0 ⇒Re(∫ ϕ(zdz)=0 ⇒  A =0 .
letA=0πcos2x2+3sin(2x)dxA=0π1+cos(2x)2(2+3sin(2x))dx=2x=t02π1+cos(t)2(2+3sin(t))dt2=1402π1+cos(t)2+3sin(t)dt4A=02π1+cost2+3sintdt=Re(02π1+eit2+3sintdt)letchangementeit=zgive02π1+eit2+3sintdt=z∣=11+z2+3zz12idziz=z∣=12i(1+z)iz(4i+3z3z1)dz=z∣=12(1+z)4iz+3z23dzletconsiderthecomplexfunctionφ(z)=2(1+z)3z2+4iz3polesofφ?letfindrootsof3z2+4iz3Δ=(2i)23(3)=4+9=5z1=2i+53andz2=2i53(arepolesofφ)z11=134+5=3>0z1isoutofcircleRes(φ,z1)=0z21=134+5=3>0z2isoutofcircleRes(φ,z2)=0z∣=1φ(z)dz=2iπ{Res(φ,z1)+Res(φ,z2)}=0Re(φ(zdz)=0A=0.
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
∫(((1+cos2x)/2)/(2+3sin2x))dx  (1/2)∫((1+cos2x)/(2+3sin2x))dx  (1/2)∫(dx/(2+3(((2tanx)/(1+tan^2 x)))))+(1/(12))∫((d(2+3sin2x))/(2+3sin2x))  (1/2)∫((sec^2 xdx)/(2+2tan^2 x+6tanx))+(1/(12))ln(2+3sin2x)+c  (1/4)∫((d(tanx))/(tan^2 x+3tanx+1))+do  (1/4)∫((d(tanx))/(tan^2 x+2tanx×(3/2)+(9/4)+1−(9/4)))+do  (1/4)∫((d(tanx))/((tanx+(3/2))^2 −((√(5/4)) )^2 ))+do  (1/4)∫((d(tanx+(3/2)))/((tanx+(3/2))^2 −((√(5/4)) )^2 ))+do  (1/4)×(1/(2(((√5)/2))))ln(((tanx+(3/2)−((√5)/2))/(tanx+(3/2)+((√5)/2))))+(1/(12))ln(2+3sin2x)+c  now we have to put limit..  (1/(4(√5)))∣ln(((tanx+(3/2)−((√5)/2))/(tanx+(3/2)+((√5)/2))))+(1/(12))ln(2+3sin2x)∣_0 ^π   answer is zero...
1+cos2x22+3sin2xdx121+cos2x2+3sin2xdx12dx2+3(2tanx1+tan2x)+112d(2+3sin2x)2+3sin2x12sec2xdx2+2tan2x+6tanx+112ln(2+3sin2x)+c14d(tanx)tan2x+3tanx+1+do14d(tanx)tan2x+2tanx×32+94+194+do14d(tanx)(tanx+32)2(54)2+do14d(tanx+32)(tanx+32)2(54)2+do14×12(52)ln(tanx+3252tanx+32+52)+112ln(2+3sin2x)+cnowwehavetoputlimit..145ln(tanx+3252tanx+32+52)+112ln(2+3sin2x)0πansweriszero

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