Question Number 102183 by bemath last updated on 07/Jul/20

∫ ((cos θ)/(sin θ−cos θ)) dθ ?

Answered by Dwaipayan Shikari last updated on 07/Jul/20

(1/2)log(sinθ−cosθ)−(θ/2)+C  I=∫((cosθ)/(sinθ−cosθ))dθ  I=∫((sinθ)/(sinθ−cosθ))−1dθ  2I=∫((sinθ+cosθ)/(sinθ−cosθ))−1dθ  I=(1/2)log(sinθ−cosθ)−(θ/2)+Constant

Answered by bobhans last updated on 07/Jul/20

((cos θ)/(sin θ−cos θ)) × ((sin θ+cos θ)/(sin θ+cos θ)) = (((1/2)sin 2θ+cos ^2 θ)/(−cos 2θ))  (1/4)∫((d(cos 2θ))/(cos 2θ))−∫((cos ^2 θ)/(cos 2θ)) dθ =   (1/4) ln∣cos 2θ∣ −∫(((1/2)+(1/2)cos 2θ)/(cos 2θ)) dθ =  (1/4)ln∣cos 2θ∣−(1/2)∫sec 2θ−(1/2)θ + c =  (1/4)ln∣cos 2θ∣−(1/4)ln∣sec 2θ+tan 2θ∣−(1/2)θ + c

Answered by 1549442205 last updated on 07/Jul/20

Putting t=tan(θ/2)⇒dt=(1/2)(1+tan^2 (θ/2))dθ=(1/2)(1+t^2 )dθ  cosθ=((1−t^2 )/(1+t^2 )), sinθ=((2t)/(1+t^2 )), dθ=((2dt)/(1+t^2 ))  F=2∫((1−t^2 )/((t^2 +2t−1)(1+t^2 )))dt==−∫((t+1)/(t^2 +1))dt+∫((t+1)/(t^2 +2t−1))dt  =−∫(dt/(1+t^2 ))−(1/2)∫((d(t^2 +1))/(t^2 +1))+(1/2)∫((d(t^2 +2t−1))/(t^2 +2t−1))  =−arctan t−(1/2)ln(1+t^2 )+(1/2)ln∣t^2 +2t−1∣  ((−𝛉)/2)−(1/2)ln(tan^2 (𝛉/2)+1)+(1/2)ln∣tan^2 (𝛉/2)+2tan(𝛉/2)−1∣+C

Answered by mathmax by abdo last updated on 08/Jul/20

I =∫ ((cosθ)/(snθ−cosθ))dθ ⇒ I =∫  (dθ/(tanθ−1)) changement tanθ =x give  I =∫  (dx/((1+x^2 )(x−1))) let decompose F(x) =(1/((x−1)(x^2  +1))) ⇒  F(x) =(a/(x−1)) +((bx+c)/(x^2  +1))  wehave a =(1/2) ,  lim_(x→+∞)  xF(x) =0 =a+b ⇒b =−(1/2)  F(0) =−1 =−a +c ⇒c =a−1 =−(1/2) ⇒F(x)=(1/(2(x−1))) +((−(1/2)x−(1/2))/(x^2  +1)) ⇒  I = (1/2)∫ (dx/(x−1))−(1/2)∫ ((x+1)/(x^2  +1))dx =(1/2)ln∣x−1∣−(1/4)ln(x^2  +1)−(1/2) arctanx +c  =(1/2)ln∣tanθ −1∣ −(1/4)ln(1+tan^2 θ)−(θ/2) +c .