Question Number 102926 by mohammad17 last updated on 11/Jul/20

 ∫ (dθ/(2sin^2 θ−cos^2 θ))  ?

Answered by OlafThorendsen last updated on 11/Jul/20

∫(dθ/(2((2t)/(1+t^2 ))−((1−t^2 )/(1+t^2 ))))  (t = tan(θ/2))  ∫((1+t^2 )/(t^2 +4t−1))dθ  ∫((1+t^2 )/((t+2)^2 −5))dθ  ∫((1+t^2 )/(2(√5)))[(1/(t+2−(√5)))−(1/(t+2+(√5)))]dθ  (1/(√5))ln((t+2−(√5))/(t+2+(√5)))+C  (1/(√5))ln((tan(θ/2)+2−(√5))/(tan(θ/2)+2+(√5)))+C

Commented bymohammad17 last updated on 11/Jul/20

thank you sir

Answered by Dwaipayan Shikari last updated on 11/Jul/20

∫((sec^2 θdθ)/(2tan^2 θ−1))=∫(dt/(2t^2 −1))=(1/2)∫(dt/(t^2 −(1/2)))=(1/(2(√2)))log((((√2)t−1)/((√2)t+1)))+C  (take  tanθ=x  (1/(2(√2)))log((((√2)tanθ−1)/((√2)tanθ+1)))+C

Commented bymohammad17 last updated on 11/Jul/20

very nice thank you sir

Answered by mathmax by abdo last updated on 11/Jul/20

I =∫  (dx/(2sin^2 x−cos^2 x)) ⇒ I =∫  (dx/(2sin^2 x−(1−sin^2 x)))  =∫  (dx/(2sin^2 x−1+sin^2 x)) =∫  (dx/(3sin^2 x−1)) =∫  (dx/(((√3)sinx−1)((√3)sinx+1)))  =(1/2)∫ ((1/((√3)sinx−1))−(1/((√3)sinx +1)))dx =H−K  H =(1/2) ∫ (dx/((√3)sinx −1)) ⇒2H =_(tan((x/2))=t)     ∫   ((2dt)/((1+t^2 )((√3)×((2t)/(1+t^2 )) −1)))  =∫  ((2dt)/(2(√3)t−1−t^2 )) =−2 ∫  (dt/(t^2 −2(√3)t +1))  t^2 −2(√3)t +1 =0 →Δ^′  =3−1=2 ⇒t_1 =(√3)+(√2)  and t_2 =(√3)−(√2)  2H =−2 ∫  (dt/((t−t_1 )(t−t_2 ))) =((−1)/(√2)) ∫((1/(t−t_1 ))−(1/(t−t_2 )))dt  H=−(1/(2(√2)))ln∣((t−t_1 )/(t−t_2 ))∣ +C =−(1/(2(√2)))ln∣((tan((x/2))−(√3)−(√2))/(tan((x/2))−(√3)+(√2)))∣ +C  K =(1/2)∫ (dx/((√3)sinx +1)) ⇒2K =_(tan((x/2))=t)    ∫ ((2dt)/((1+t^2 )((√3)×((2t)/(1+t^2 )) +1)))  ⇒K =∫  (dt/(2(√3)t +1+t^2 )) =∫  (dt/(t^2 +2(√3)t +1))  Δ^′  =2 ⇒ t_1 =−(√3)+(√2) and t_2 =−(√3)−(√2) ⇒  K =∫ (dt/((t−t_1 )(t−t_2 ))) =(1/(2(√2)))∫((1/(t−t_1 ))−(1/(t−t_2 )))dt  =(1/(2(√2)))ln∣((t−t_1 )/(t−t_2 ))∣ +C =(1/(2(√2)))ln∣((tan((x/2))+(√3)−(√2))/(tan((x/2))+(√3)+(√2)))∣ +C  I =H−K