Question Number 103345 by harckinwunmy last updated on 14/Jul/20

Commented byharckinwunmy last updated on 14/Jul/20

please help

Commented byDwaipayan Shikari last updated on 14/Jul/20

2∫_0 ^0 (((2t)/(1+t^2 ))/((5/4)−((1−t^2 )/(1+t^2 )))) .(1/(1+t^2 ))dt=0 {take tan(θ/2)=t

Commented bymathmax by abdo last updated on 14/Jul/20

not correct sir you must have a bijection...!

Commented byDwaipayan Shikari last updated on 14/Jul/20

∫_0 ^0 f(x)dx=0

Commented bymathmax by abdo last updated on 15/Jul/20

tan is bijective on ]−(π/2),(π/2)[ not [0,2π] be carefull

Answered by abdomathmax last updated on 14/Jul/20

Q2 (∂u/∂x)(x,y) =((1(x^2 +y^2 )−2x(x−y))/((x^2 +y^2 )^2 )) =((−x^2 +y^2 +2xy)/((x^2 +y^2 )^2 )) (∂^2 u/∂x^2 )(x,y) =(((−2x+2y)(x^2 +y^2 )^2 −2(2x)(x^2 +y^2 ))/((x^2 +y^2 )^4 )) =(((−2x+2y)(x^2 +y^2 )−4x)/((x^2 +y^2 )^3 )) =((−2x^3 −2xy^2 +2yx^2 −4x)/((x^2 +y^2 )^3 )) after we calculate (∂^2 u/∂y^2 )(x,y) and prove that (∂^2 u/∂x^2 ) +(∂^2 u/∂y^2 ) =0

Answered by abdomathmax last updated on 14/Jul/20

let I =∫_0 ^5 (25−t^2 )^(3/2) dt we do the changeme7t t =5x ⇒ I =∫_0 ^1 (25−25x^2 )^(3/2) 5dx =5.(5^2 )^(3/(2 )) ∫_0 ^1 (1−x^2 )^(3/2) dx =_(x =u^(1/2) ) 5^(4 ) ∫_0 ^1 (1−u)^(3/2) (1/2)u^(−(1/2)) du =(5^4 /2) ∫_0 ^1 u^(−(1/2)) (1−u)^(3/2) du we havd B(p,q) =∫_0 ^1 x^(p−1) (1−x)^(q−1) dx p−1 =−(1/2) ⇒p =(1/2) and q−1 =(3/2) ⇒q =1+(3/2)=(5/2) ⇒ I =(5^4 /2) ×B((1/2),(5/2)) .

Answered by abdomathmax last updated on 14/Jul/20

A =∫_0 ^(2π) ((sinθ)/((5/4)−cosθ))dθ ⇒A =4∫_0 ^(2π ) ((sinθ)/(5−4cosθ))dθ changement e^(iθ) =z give A =4 ∫_(∣z∣=1) (((z−z^(−1) )/(2i))/(5−4×((z+z^(−1) )/2)))×(dz/(iz)) =−2 ∫_(∣z∣=1) ((2(z−z^(−1) ))/(z(10−4z−4z^(−1) )))dz =−4 ∫_(∣z∣=1) ((z−z^(−1) )/(10z−4z^2 −4z))dz =4 ∫_(∣z∣=1) ((z^2 −1)/(z(4z^2 −10z +4)))dz =2 ∫_(∣z∣=1) ((z^2 −1)/(z(2z^2 −5z +2)))dz let ϕ(z) =((z^2 −1)/(z{2z^2 −5z +2})) poles of ϕ? 2z^2 −5z +2 =0 →Δ =25−16 =9 ⇒ z_1 =((5+3)/4) =2 and z_2 =((5−3)/4) =(1/2) ∫_(∣z∣ =1) ϕ(z)dz =2iπ {Res(ϕ,0) +Res(ϕ,z_2 )} ϕ(z) =((z^2 −1)/(2z(z−2)(z−(1/2)))) Res(ϕ,0) =lim_(z→0) zϕ(z) =−(1/2) Res(ϕ,(1/2)) =lim_(z→(1/2)) (z−(1/2))ϕ(z) =(((1/4)−1)/(2×(1/2)((1/2)−2))) =−(3/4)×(1/(−(3/2))) =(1/2) ⇒ ∫_(∣z∣=1) ϕ(z)dz =0 ⇒ A =0

Answered by 1549442205 last updated on 15/Jul/20

Q3.F=∫((sinθ)/((5/4)−cosθ))dθ=−∫((dcosθ)/((5/4)−cosθ))=_(u=cos𝛉) ∫(du/(u−(5/4))) .∫((d(u−(5/4)))/(u−(5/4)))=ln∣u−(5/4)∣=ln((5/4)−cosθ)∣_0 ^(2π) =ln((1/4))−ln((1/4))=0 Q2.b/∫_0 ^5 (25−t^2 )^(3/2) dt=F.Put t=5sinθ ⇒dt=5cosθdθ.Then F=∫_0 ^(π/2) (25cos^2 θ)^(3/2) 5cosθdθ=625∫_0 ^(π/2) (cosθ)^4 dθ =((625)/2)∫_0 ^(π/2) 2(sinθ)^(2.(1/2)−1) (cosθ)^(2.(5/2)−1) dθ =((625)/2)B((5/2),(1/2))=((625)/2)×((Γ((5/2)).Γ((1/2)))/(Γ((5/2)+(1/2)))) =((625)/2)×((((3(√π))/4)×(√π))/(Γ(3)))=((625π)/8)×(3/2)(as Γ(3)=2!) Thus,F=((1875𝛑)/(16))