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clculate-0-1-x-x-2-2x-2-dx-




Question Number 31069 by abdo imad last updated on 02/Mar/18
clculate  ∫_0 ^1  x(√(x^2  −2x+2)) dx
clculate01xx22x+2dx
Commented by abdo imad last updated on 03/Mar/18
let put I=∫_0 ^1  x(√(x^2  −2x +2dx))   I= ∫_0 ^1  x(√((x−1)^2  +1)) dx the ch. x−1=sht  I= ∫_(argsh(−1)) ^o  (1+sht)cht chtdt=−∫_0 ^(ln(−1+(√2))) ch^2 tdt  −∫_0 ^(ln(−1+(√2))) sht ch^2 tdt  but wehave  ∫_0 ^(ln(−1+(√2))) ch^2 tdt= ∫_0 ^(ln(−1+(√2))) ((1+ch(2t))/2)dt  =(1/2)ln(−1+(√2)) +(1/4)[sh(2t)]_0 ^(ln(−1+(√2) ))   =(1/2)ln(−1+(√2)) +(1/4)sh(2ln(−1+(√2))) and  ∫_0 ^(ln(−1+(√2))) sht ch^2 t dt=(1/3)[ ch^3 t]_0 ^(ln(−1+(√2)))   =(1/3) (ch^3 (ln(−1+(√2))−1) by using shu =((e^u  −e^(−u) )/2) and  chu= ((e^u  +e^(−u) )/2) tbe value of I isdetermined.
letputI=01xx22x+2dxI=01x(x1)2+1dxthech.x1=shtI=argsh(1)o(1+sht)chtchtdt=0ln(1+2)ch2tdt0ln(1+2)shtch2tdtbutwehave0ln(1+2)ch2tdt=0ln(1+2)1+ch(2t)2dt=12ln(1+2)+14[sh(2t)]0ln(1+2)=12ln(1+2)+14sh(2ln(1+2))and0ln(1+2)shtch2tdt=13[ch3t]0ln(1+2)=13(ch3(ln(1+2)1)byusingshu=eueu2andchu=eu+eu2tbevalueofIisdetermined.
Answered by Joel578 last updated on 02/Mar/18
I = ∫_0 ^1  (x − 1 + 1)(√(x^2  − 2x + 2)) dx     = ∫_0 ^1  (x − 1)(√(x^2  − 2x + 2)) dx + ∫_0 ^1  (√((x −1)^2  + 1)) dx    I_1  = ∫ (x − 1)(√(x^2  − 2x + 2)) dx  u = x^2  − 2x + 2  →  du = 2(x −1) dx  I_1  = (1/2)∫ (x − 1)(√u) . (du/(x − 1))       = (1/3)u(√u) = (1/3)(x^2  − 2x + 2)^(3/2)  + C    I_2  = ∫ (√((x − 1)^2  + 1)) dx  x − 1 = tan θ  →  dx = sec^2  θ dθ  I_2  = ∫ (√(tan^2  θ + 1)) . sec^2  θ dθ       = ∫ sec^3  θ dθ       = ((tan θ sec θ)/2) − (1/2)ln ∣sec θ + tan θ∣ + C       = (((x − 1)(√(x^2  − 2x + 2)))/2) − (1/2)ln ∣(√(x^2  − 2x + 2)) + x − 1∣ + C    I = [I_1  + I_2 ]_0 ^1      = [(1/3)(x^2  − 2x + 2)^(3/2)  + (((x − 1)(√(x^2  − 2x + 2)))/2) − (1/2)ln ∣(√(x^2  − 2x + 2)) + x − 1∣]_0 ^1      = ((1/3) + 0 − 0) − (((2(√2))/3) − ((√2)/2) − (1/2)ln ((√2) − 1))     = ((1 − 2(√2))/3) + (1/2)((√2) + ln ((√2) − 1))
I=01(x1+1)x22x+2dx=01(x1)x22x+2dx+01(x1)2+1dxI1=(x1)x22x+2dxu=x22x+2du=2(x1)dxI1=12(x1)u.dux1=13uu=13(x22x+2)32+CI2=(x1)2+1dxx1=tanθdx=sec2θdθI2=tan2θ+1.sec2θdθ=sec3θdθ=tanθsecθ212lnsecθ+tanθ+C=(x1)x22x+2212lnx22x+2+x1+CI=[I1+I2]01=[13(x22x+2)32+(x1)x22x+2212lnx22x+2+x1]01=(13+00)(2232212ln(21))=1223+12(2+ln(21))

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