Question Number 143399 by Ar Brandon last updated on 13/Jun/21

Answered by Ar Brandon last updated on 13/Jun/21
![I=∫_0 ^(2π) (dx/((1−ksinx)^2 ))=2∫_(−(π/2)) ^(π/2) (dx/((1−ksinx)^2 )) t=tan(x/2) =2∫_(−1) ^1 (1/((1−k((2t)/(1+t^2 )))^2 ))∙((2dt)/(1+t^2 ))=4∫_(−1) ^1 ((t^2 +1)/((t^2 −2kt+1)^2 ))dt =4[((at+b)/(t^2 −2kt+1))+∫((ct+d)/(t^2 −2kt+1))dt]_(−1) ^1 =4∫[((a(t^2 −2kt+1)−(at+b)(2t−2k))/((t^2 +2kt+1)^2 ))+((ct+d)/(t^2 −2kt+1))]dt c=0,a−2a+d=1⇒d=a+1,−2ka+2ka−2b−2kd=0⇒b=−kd a+2bk+d=1⇒d+bk=1⇒d(1−k^2 )=1⇒d=(1/(1−k^2 )),b=(k/(k^2 −1)) ⇒a=(1/(1−k^2 ))−1=((1+k^2 −1)/(1−k^2 ))=(k^2 /(1−k^2 )) I=4[(((k^2 /(1−k^2 ))t+(k/(k^2 −1)))/(t^2 −2kt+1))+(1/(1−k^2 ))∫(dt/(t^2 −2kt+1))]_(−1) ^1 =(4/(2−2k))((k^2 /(1−k^2 ))+(1/(k^2 −1)))−(4/(2+2k))((k/(k^2 −1))−(k^2 /(1−k^2 )))+(4/(1−k^2 ))∫(dt/(t^2 −2kt+1)) ∫(dt/(t^2 −2kt+1))=∫(dt/((t−k)^2 +1−k^2 ))=(1/( (√(1−k^2 ))))arctan(((t−k)/( (√(1−k^2 ))))), ∣k∣<1 ∫(dt/(t^2 −2kt+1))=∫(dt/((t−k)^2 −(k^2 −1)))=((−1)/( (√(k^2 −1))))argtanh(((t−k)/( (√(k^2 −1))))), ∣k∣>1 I=(2/(k−1))−(2/(k+1))((k/(k−1)))+((4(√(1−k^2 )))/( (1−k^2 )^2 ))[arctan((√((1−k)/(1+k))))+arctan(((k+1)/( (√(1−k^2 )))))], ∣k∣<1 =(2/(k−1))−(2/(k+1))((k/(k−1)))+((2(√(k^2 −1)))/((k^2 −1)^2 ))ln∣(((√(k^2 −1))−k+1)/( (√(k^2 −1))+k−1))∣, ∣k∣>1](https://www.tinkutara.com/question/Q143400.png)
Commented by Ar Brandon last updated on 13/Jun/21

Answered by mathmax by abdo last updated on 13/Jun/21

Commented by Ar Brandon last updated on 13/Jun/21

Commented by mathmax by abdo last updated on 13/Jun/21
