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solve-I-0-1-Arcsin-x-1-x-x-2-dx-




Question Number 150356 by mnjuly1970 last updated on 11/Aug/21
 solve...       I:= ∫_0 ^( 1) (( Arcsin ((√x) ))/(1−x + x^( 2) )) dx=?
solveI:=01Arcsin(x)1x+x2dx=?
Answered by Lordose last updated on 12/Aug/21
  Ω = ∫_0 ^( 1) ((sin^(−1) ((√x)))/(1−x+x^2 ))dx =^(x=u^2 ) 2∫_0 ^( 1) ((usin^(−1) (u))/(1−u^2 +u^4 ))du  Ω =  2∫_0 ^( 1) ((usin^(−1) (u))/(1−u^2 +u^4 ))du =^(u=sin(y)) ∫_0 ^( (𝛑/2)) ((ysin(2y))/(cos^2 (y)+sin^4 (y)))dy  Ω =^(y=(x/2)) 2∫_0 ^( 𝛑) ((xsin(x))/(cos^2 (x)+3))dx   Ω = (1/2)∫_0 ^( 𝛑) ((xsin(x))/(cos^2 (x)+3))dx = (1/2)∫_0 ^( 𝛑) (((𝛑−x)sin(x))/(cos^2 (x)+3))dx  Ω = (𝛑/2)∫_0 ^( 𝛑) ((sin(x))/(cos^2 (x)+3))dx =^(u=cos(x)) (𝛑/2)∫_(−1) ^( 1) (1/(u^2 +3))du  Ω = ∣(𝛑/( 2(√3)))tan^(−1) ((u/( (√3))))∣_(−(𝛑/4)) ^(𝛑/4) = (𝛑^2 /( 6(√3)))
Ω=01sin1(x)1x+x2dx=x=u2201usin1(u)1u2+u4duΩ=201usin1(u)1u2+u4du=u=sin(y)0π2ysin(2y)cos2(y)+sin4(y)dyΩ=y=x220πxsin(x)cos2(x)+3dxΩ=120πxsin(x)cos2(x)+3dx=120π(πx)sin(x)cos2(x)+3dxΩ=π20πsin(x)cos2(x)+3dx=u=cos(x)π2111u2+3duΩ=π23tan1(u3)π4π4=π263
Answered by mindispower last updated on 11/Aug/21
f(x)=1−x+x^2   f(1−x)=1−(1−x)+(1−x)^2 =1−x+x^2   x→1−x  I=∫_0 ^1 ((arcsin(√((1−x))))/(1−x+x^2 ))dx  arcsin(a)+arcsin(b)=arcsin(a(√(1−b^2 ))+b(√(1−a^2 )))  arcsin((√(1−x)))+arcsin((√x))=arcsin((1−x)(√(1−(√x^2 )+(√x).(√(1−((√(1−x)^2 )))))))  x∈[0,1[ (√((1−x)^2 ))=1−x  =arcsin(1−x+x)=arcsin(1)=(π/2)  2I=∫_0 ^1 ((arcsin((√(1−x))))/(1−x+x^2 ))dx+∫_0 ^1 ((arcsin((√x)))/(1−x+x^2 ))dx  2I=∫_0 ^1 ((arcsin((√x))+arcsim((√(1−x))))/(1−x+x^2 ))dx  =(π/2)∫_0 ^1 (dx/(1−x+x^2 ))=(π/2)∫_0 ^1 (dx/((x−(1/2))^2 +(3/4)))  =(π/2)[(2/( (√3)))arctan((2/( (√3)))(x−(1/2))]_0 ^1   =(π/( (√3))).(π/3)=((2π^2 )/(3(√3)))  I=(π^2 /(3(√3)))
f(x)=1x+x2f(1x)=1(1x)+(1x)2=1x+x2x1xI=01arcsin(1x)1x+x2dxarcsin(a)+arcsin(b)=arcsin(a1b2+b1a2)arcsin(1x)+arcsin(x)=arcsin((1x)1x2+x.1(1x)2)x[0,1[(1x)2=1x=arcsin(1x+x)=arcsin(1)=π22I=01arcsin(1x)1x+x2dx+01arcsin(x)1x+x2dx2I=01arcsin(x)+arcsim(1x)1x+x2dx=π201dx1x+x2=π201dx(x12)2+34=π2[23arctan(23(x12)]01=π3.π3=2π233I=π233
Answered by Kamel last updated on 11/Aug/21
     I=2∫_0 ^1 ((tArcsin(t))/(1−t^2 +t^4 ))dt=2∫_0 ^(π/2) ((usin(u)cos(u)du)/(cos^2 (u)+sin^4 (u)))       =∫_0 ^π ((usin(u)du)/(3+cos^2 (u)))=∫_0 ^π (((π−t)sin(t))/(3+cos^2 (t)))dt  ∴ I=(π/2)∫_0 ^π ((sin(u)dt)/(3+cos^2 (t)))=π∫_0 ^(π/2) ((sin(t)dt)/(3+cos^2 (t)))=^(y=cos(t)) π∫_0 ^1 (dy/(3+y^2 ))=(π/( (√3)))Arctg((1/( (√3))))        =(π^2 /(6(√3)))
I=201tArcsin(t)1t2+t4dt=20π2usin(u)cos(u)ducos2(u)+sin4(u)=0πusin(u)du3+cos2(u)=0π(πt)sin(t)3+cos2(t)dtI=π20πsin(u)dt3+cos2(t)=π0π2sin(t)dt3+cos2(t)=y=cos(t)π01dy3+y2=π3Arctg(13)=π263

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