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Question Number 107184 by bemath last updated on 09/Aug/20
@bemath@ ∫_0 ^2 x ((8−x^3 ))^(1/3) dx =?
@bemath@∫20x8−x33dx=?
Answered by john santu last updated on 09/Aug/20
Commented by john santu last updated on 09/Aug/20
typo 8v = x^3
typo8v=x3
Commented by bobhans last updated on 09/Aug/20
typo the last line : ((16π)/(9(√3)))
typothelastline:16π93
Commented by john santu last updated on 09/Aug/20
haha...thanks
haha…thanks
Answered by 1549442205PVT last updated on 12/Aug/20
Set (8/x^3 )−1=t^3 ⇒3t^2 dt=−((24)/x^4 )dx ⇒^3 (√(8−x^3 )) =xt,dx=−((x^4 t^2 dt)/8) ⇒F=∫((x^2 t.x^4 t^2 )/(−8))dt=−(1/8)∫ t^3 x^6 dt =((−1)/8)∫t^3 .((8/(1+t^3 )))^2 dt=−8∫(t^3 /((1+t^3 )^2 ))dt =−8∫((t^3 +1−1)/((1+t^3 )^2 ))dt=−8∫(dt/(1+t^3 ))−8∫(dt/((1+t^3 )^2 )) (1/(1+t^3 ))=(a/(1+t))+((bt+c)/(t^2 −t+1))⇔(((a+b)t^2 +(b+c−a)t+a+c)/((1+t)(t^2 −t+1)))=(1/(t^3 +1)) ⇔ { ((a+b=0)),((b+c−a=0)),((a+c=1)) :} ⇔ { ((a=1/3)),((b=−1/3)),((c=2/3)) :} A=∫(dt/(1+t^3 ))=∫(dt/(3(1+t)))−∫(((t−2)dt)/(3(t^2 −t+1))) =(1/3)lnt−(1/6)∫((2t−1−3)/((t^2 −t+1)))dt=(1/3)lnt −(1/6)∫((d(t^2 −t+1))/(t^2 −t+1))+(1/2)∫(dt/((t−(1/2))^2 +(((√3)/2))^2 )) =(1/3)lnt−(1/6)ln(t^2 −t+1)+(1/2)×(2/( (√3)))tan^(−1) (((2x−1)/( (√3)))) B=(1/((1+t^3 )^2 ))=[(1/3)((1/(t+1))−((t−2)/(t^2 −t+1)))]^2 9B=(1/((1+t)^2 ))−((t^2 −4t+4)/((t^2 −t+1)^2 ))−((2(t−2))/((t+1)(t^2 −t+1))) =(1/((1+t)^2 ))−((t^2 −t+1−3t+3)/((t^2 −t+1)^2 ))−(2/(t^2 −t+1))+2((1/(t+1))−((t−2)/(t^2 −t+1))) =(1/((1+t)^2 ))−(1/(t^2 −t+1))+((3(t−1))/((t^2 −t+1)^2 ))−(2/(t^2 −t+1))+(2/(t+1))−((2(t−2))/(t^2 −t+1)) =(1/((t+1)^2 ))+(2/(t+1))+((3(t−1))/((t^2 −t+1)^2 ))−((2t−1)/(t^2 −t+1)) ∫(dt/((1+t^3 )^2 ))=∫((1/((t+1)^2 ))+(2/(t+1))+((3(t−1))/((t^2 −t+1)^2 ))−((2t−1)/(t^2 −t+1)))dt =∫((d(t+1))/((t+1)^2 ))+2∫((d(t+1))/(t+1))+(3/2)∫((2t−1)/((t^2 −t+1)^2 ))dt−(3/2)∫(dt/((t^2 −t+1)^2 ))−∫ ((d(t^2 −t+1))/(t^2 −t+1)) =−(1/(t+1))+2ln(t+1)−(3/(2(t^2 −t+1)))−ln(t^2 −t+1)−(3/(2(t^2 −t+1)^2 )) I_2 =∫(dt/((t^2 −t+1)^2 ))=∫((d(t−(1/2)))/([(t−(1/2))^2 +(((√3)/2))^2 ]^2 )).Set u=t−(1/2),a=((√3)/2) I_2 =∫(du/((u^2 +a)^2 ))=(1/(2a^2 (2−1))).(u/((u^2 +a^2 )^(2−1) ))+(1/a^2 ).((2.2−3)/(2.2−2)).I_1 =(2/3).(u/((u^2 +a^2 )))+(4/3).(1/2)∫(du/(u^2 +a^2 )) =(2/3).((t−(1/2))/((t^2 −t+1)))+(2/3).(2/( (√3)))tan^(−1) (((2t−1)/( (√3)))) =((2t−1)/(3(t^2 −t+1)))+(4/(3(√3)))tan^(−1) (((2t−1)/( (√3)))) Hence, 9B=((−1)/(t+1))+2ln(t+1)−(3/(2(t^2 −t+1)))−ln(t^2 −t+1)−(3/(2(t^2 −t+1)^2 )) =((−1)/(t+1))−(3/(2(t^2 −t+1)))+2ln((t+1)/(t^2 −t+1))−3I_2 =((−1)/(t+1))−(3/(2(t^2 −t+1)))+2ln((t+1)/(t^2 −t+1))−((2t−1)/(t^2 −t+1))−(4/( (√3)))tan^(−1) (((2t−1)/( (√3)))) F=−8A−8B=−8[(1/3)lnt−(1/6)ln(t^2 −t+1)+(1/2)×(2/( (√3)))tan^(−1) (((2x−1)/( (√3))))] −(8/9)[=((−1)/(t+1))−(3/(2(t^2 −t+1)))+2ln((t+1)/(t^2 −t+1))−((2t−1)/(t^2 −t+1))−(4/( (√3)))tan^(−1) (((2t−1)/( (√3))))]
Set8x3−1=t3⇒3t2dt=−24x4dx⇒38−x3=xt,dx=−x4t2dt8⇒F=∫x2t.x4t2−8dt=−18∫t3x6dt=−18∫t3.(81+t3)2dt=−8∫t3(1+t3)2dt=−8∫t3+1−1(1+t3)2dt=−8∫dt1+t3−8∫dt(1+t3)211+t3=a1+t+bt+ct2−t+1⇔(a+b)t2+(b+c−a)t+a+c(1+t)(t2−t+1)=1t3+1⇔{a+b=0b+c−a=0a+c=1⇔{a=1/3b=−1/3c=2/3A=∫dt1+t3=∫dt3(1+t)−∫(t−2)dt3(t2−t+1)=13lnt−16∫2t−1−3(t2−t+1)dt=13lnt−16∫d(t2−t+1)t2−t+1+12∫dt(t−12)2+(32)2=13lnt−16ln(t2−t+1)+12×23tan−1(2x−13)B=1(1+t3)2=[13(1t+1−t−2t2−t+1)]29B=1(1+t)2−t2−4t+4(t2−t+1)2−2(t−2)(t+1)(t2−t+1)=1(1+t)2−t2−t+1−3t+3(t2−t+1)2−2t2−t+1+2(1t+1−t−2t2−t+1)=1(1+t)2−1t2−t+1+3(t−1)(t2−t+1)2−2t2−t+1+2t+1−2(t−2)t2−t+1=1(t+1)2+2t+1+3(t−1)(t2−t+1)2−2t−1t2−t+1∫dt(1+t3)2=∫(1(t+1)2+2t+1+3(t−1)(t2−t+1)2−2t−1t2−t+1)dt=∫d(t+1)(t+1)2+2∫d(t+1)t+1+32∫2t−1(t2−t+1)2dt−32∫dt(t2−t+1)2−∫d(t2−t+1)t2−t+1=−1t+1+2ln(t+1)−32(t2−t+1)−ln(t2−t+1)−32(t2−t+1)2I2=∫dt(t2−t+1)2=∫d(t−12)[(t−12)2+(32)2]2.Setu=t−12,a=32I2=∫du(u2+a)2=12a2(2−1).u(u2+a2)2−1+1a2.2.2−32.2−2.I1=23.u(u2+a2)+43.12∫duu2+a2=23.t−12(t2−t+1)+23.23tan−1(2t−13)=2t−13(t2−t+1)+433tan−1(2t−13)Hence,9B=−1t+1+2ln(t+1)−32(t2−t+1)−ln(t2−t+1)−32(t2−t+1)2=−1t+1−32(t2−t+1)+2lnt+1t2−t+1−3I2=−1t+1−32(t2−t+1)+2lnt+1t2−t+1−2t−1t2−t+1−43tan−1(2t−13)F=−8A−8B=−8[13lnt−16ln(t2−t+1)+12×23tan−1(2x−13)]−89[=−1t+1−32(t2−t+1)+2lnt+1t2−t+1−2t−1t2−t+1−43tan−1(2t−13)]

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