Question and Answers Forum
Question Number 51824 by Abdo msup. last updated on 30/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18
![∫(dx/(cos2x+2cos2x.cosx)) ∫(dx/(cos2x(1+2cosx))) ∫(dx/((2cos^2 x−1)(2cosx+1))) cosx=((1−t^2 )/(1+t^2 )) t=tan(x/2) (dt/dx)=(1/2)sec^2 (x/2) so ((2dt)/(1+t^2 ))=dx ∫(((2dt)/(1+t^2 ))/([2(((1−t^2 )/(1+t^2 )))^2 −1][2(((1−t^2 )/(1+t^2 )))+1])) ∫(((2dt)/(1+t^2 ))/([((2−4t^2 +2t^4 −1−2t^2 −t^4 )/((1+t^2 )^2 ))][((2−2t^2 +1+t^2 )/(1+t^2 ))])) ∫((((2(1+t^2 )^3 )/(1+t^2 ))dt)/((t^4 −6t^2 +1)(3−t^2 ))) 2∫(((1+t^2 )^2 )/((t^4 −6t^2 +1)(3−t^2 )))dt 2∫((t^4 +2t^2 +1)/((t^4 −6t^2 +1)(3−t^2 )))dt now t^4 +2t^2 +1=a(t^4 −6t^2 +1)+(bt^2 +c)(3−t^2 ) t^4 +2t^2 +1=at^4 −6at^2 +a+3bt^2 −bt^4 +3c−ct^2 t^4 +2t^2 +1=t^4 (a−b)+t^2 (−6a+3b−c)+(a+c) a−b=1 −6a+3b−c=2 a+c=1 −6a+3(a−1)−(1−a)=2 −6a+3a+a−3−1=2 −2a=6 a=−3 b=a−1=−4 c=1−a=4 a=−3 b=−4 c=4 2∫((a(t^4 −6t^2 +1)+(bt^2 +c)(3−t^2 ))/((t^4 −6t^2 +1)(3−t^2 )))dt 2∫((−3)/(3−t^2 ))dt+2∫(((−4t^2 +4))/(t^4 −6t^2 +1))dt −6∫(dt/(3−t^2 ))−8∫((t^2 −1)/(t^4 −6t^2 +1))dt 6∫(dt/(t^2 −3))−8∫((1−(1/t^2 ))/((t^2 +(1/t^2 ))−6)) 6∫(dt/(t^2 −3))−8∫((d(t+(1/t)))/((t+(1/t))^2 −8)) 6×(1/(2(√3)))ln(((t−(√3))/(t+(√3))))−8×(1/(2(√8) ))ln(((t+(1/t)−(√8))/(t+(1/t)+(√8))))+c =(√3) ln(((tan(x/2)−(√3))/(tan(x/2)+(√3))))−(√2) ln(((tan(x/2)+(1/(tan(x/2)))−2(√2))/(tan(x/2)+(1/(tan(x/2)))+2(√2))))+c](Q51836.png)
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Question and Answers Forum | ||
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Question Number 51824 by Abdo msup. last updated on 30/Dec/18 | ||
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Answered by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18 | ||
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