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Question Number 62419 by mathmax by abdo last updated on 20/Jun/19
calculate ∫_0 ^1 (2x^2 −1)(√(x^2 −2x+5))dx
calculate∫01(2x2−1)x2−2x+5dx
Commented by mathmax by abdo last updated on 21/Jun/19
let I =∫_0 ^1 (2x^2 −1)(√(x^2 −2x+5))dx we have x^2 −2x+5 =x^2 −2x+1+4 =(x−1)^2 +4 let use the changement (x−1) =2sh(t) ⇒x =2sh(t)+1 I = ∫_(argsh(−(1/2))) ^0 {2 (2sh(t)+1)^2 −1}2 ch(t) 2ch(t)dt =4 ∫_(ln(−(1/2)+(√(5/4)))) ^0 {2(4sh^2 t +4sht +1)−1}ch^2 (t)dt =4∫_(ln(((−1+(√5))/2))) ^0 { 8sh^2 t+8sh(t)+1}ch^2 t dt =32 ∫_(ln(((−1+(√5))/2))) ^0 sh^2 t ch^2 t dt +32 ∫_(ln(((−1+(√5))/2))) ^0 sh(t)ch^2 (t)dt +4 ∫_(ln(((−1+(√5))/2))) ^0 ch^2 (t)dt ∫_(ln(((−1+(√5))/2))) ^0 (shtcht)^2 dt =(1/4) ∫_(ln(((−1+(√5))/2))) ^0 sh(2t)dt =(1/8)[ch(2t)]_(ln(((−1+(√5))/2))) ^0 =(1/(16))[ e^(2t) +e^(−2t) ]_(ln(((−1+(√5))/2))) ^0 =(1/(16)){ 2 −(((−1+(√5))/2))^2 −(1/((((−1+(√5))/2))^2 ))} ∫_(ln(((−1+(√5))/2))) ^0 sh(t)ch^2 (t)dt =[(1/3)ch^3 t]_(ln(((−1+(√5))/2))) ^0 =(1/3)[ (((e^t +e^(−t) )/2))^3 ]_(ln(((−1+(√5))/2))) ^0 =(1/(24)){ 8 − { (((−1+(√5))/2))−(1/((−1+(√5))/2))}^3 } . ∫_(ln(((−1+(√5))/2))) ^0 ch^2 t dt =(1/2) ∫_(ln(((−1+(√5))/2))) ^0 (1+ch(2t))dt =−(1/2)ln(((−1+(√5))/2)) +(1/4)[sh(2t)]_(ln(((−1+(√5))/2))) ^0 =−(1/2)ln(((−1+(√5))/2))+(1/8)[ e^(2t) −e^(−2t) ]_(ln(((−1+(√5))/2))) ^0 =−(1/2)ln(((−1+(√5))/2)) +(1/8){−(((−1+(√5))/2))^2 +(1/((((−1+(√5))/2))^2 ))} the value of I is known .
letI=∫01(2x2−1)x2−2x+5dxwehavex2−2x+5=x2−2x+1+4=(x−1)2+4letusethechangement(x−1)=2sh(t)⇒x=2sh(t)+1I=∫argsh(−12)0{2(2sh(t)+1)2−1}2ch(t)2ch(t)dt=4∫ln(−12+54)0{2(4sh2t+4sht+1)−1}ch2(t)dt=4∫ln(−1+52)0{8sh2t+8sh(t)+1}ch2tdt=32∫ln(−1+52)0sh2tch2tdt+32∫ln(−1+52)0sh(t)ch2(t)dt+4∫ln(−1+52)0ch2(t)dt∫ln(−1+52)0(shtcht)2dt=14∫ln(−1+52)0sh(2t)dt=18[ch(2t)]ln(−1+52)0=116[e2t+e−2t]ln(−1+52)0=116{2−(−1+52)2−1(−1+52)2}∫ln(−1+52)0sh(t)ch2(t)dt=[13ch3t]ln(−1+52)0=13[(et+e−t2)3]ln(−1+52)0=124{8−{(−1+52)−1−1+52}3}.∫ln(−1+52)0ch2tdt=12∫ln(−1+52)0(1+ch(2t))dt=−12ln(−1+52)+14[sh(2t)]ln(−1+52)0=−12ln(−1+52)+18[e2t−e−2t]ln(−1+52)0=−12ln(−1+52)+18{−(−1+52)2+1(−1+52)2}thevalueofIisknown.

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