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Question Number 83603 by jagoll last updated on 04/Mar/20

∫ (dx/(1−2cos x))

dx12cosx

Commented by turbo msup by abdo last updated on 04/Mar/20

we use the changement tan((x/2))=t  ⇒∫  (dx/(1−2cosx)) =∫  (1/(1−2((1−t^2 )/(1+t^2 ))))×((2dt)/(1+t^2 ))  =2∫   (dt/(1+t^2 −2+2t^2 )) =2∫  (dt/(3t^2 −1))  =2 ∫  (dt/(((√3)t−1)((√3)+1)))  =∫((1/((√3)t−1))−(1/((√3)t +1)))dt  =(1/(√3))ln∣(√3)t−1∣−(1/(√3))ln∣(√3)t+1∣ +c  =(1/(√3))ln∣(((√3)t−1)/((√3)t+1))∣ +c  =(1/(√3))ln∣(((√3)tan((x/2))−1)/((√3)tan((x/2))+1))∣ +C

weusethechangementtan(x2)=tdx12cosx=1121t21+t2×2dt1+t2=2dt1+t22+2t2=2dt3t21=2dt(3t1)(3+1)=(13t113t+1)dt=13ln3t113ln3t+1+c=13ln3t13t+1+c=13ln3tan(x2)13tan(x2)+1+C

Commented by jagoll last updated on 04/Mar/20

thank you

thankyou

Commented by mathmax by abdo last updated on 04/Mar/20

you are welcome

youarewelcome

Commented by niroj last updated on 05/Mar/20

  ∫ ((  dx)/(1−2cos x))   let , cos x= ((1−tan^2 (x/2))/(1+tan^2 (x/2)))   =  ∫  (1/(1−2(((1−tan^2 (x/2))/(1+tan^2 (x/2))))))dx  =    ∫(((1+tan^2 (x/2))dx)/(1+tan^2 (x/2)−2+2tan^2 (x/2)))   = ∫((sec^2 (x/2)dx)/(3tan^2 (x/2)−1))    put tan (x/2)= t          sec^2 (x/2)dx=2dt   =  ∫((2dt)/(3t^2 −1))= (2/3)∫(1/(t^2 −(1/3)))dt  = (2/3)∫ (( 1)/((t)^2 −((1/(√3)))^2 ))dt  = (2/3)[ (1/(2.(1/(√3))))log ((t−(1/(√3)))/(t+(1/(√3))))]+C  =  ((2.1)/((3.2)/(√3)))( log (( t(√3)   −1)/( t(√3) +1)))+C  =  (1/(√3))(log ((   (√3) tan(x/2)−1)/(  (√3)   tan(x/2)+1)))+C

dx12cosxlet,cosx=1tan2x21+tan2x2=112(1tan2x21+tan2x2)dx=(1+tan2x2)dx1+tan2x22+2tan2x2=sec2x2dx3tan2x21puttanx2=tsec2x2dx=2dt=2dt3t21=231t213dt=231(t)2(13)2dt=23[12.13logt13t+13]+C=2.13.23(logt31t3+1)+C=13(log3tanx213tanx2+1)+C

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