The function y=f(x) is the solution of the differential equation dydx+xyx2-1=x4+21-x2 in (-1,1) satisfying f(0)=0. Then 2∫-3/23/2f(x) dx is
(2)
dydx+xx2-1y=x4+2x1-x2
The I.F. of this differential equation is e∫xx2-1 dx=e-∫x1-x2 dx=e12log(1-x2)=1-x2
The solution is given by y1-x2=∫x3+21-x21-x2 dx+λ
=∫(x4+2x) dx+λ=x55+x2+λ
At y(0)=0⇒λ=0⇒y1-x2=x55+x2
∫-3/23/2x55+x21-x2 dx=∫-3/23/2x21-x2 dx
(The other part is odd) =2∫03/2x21-x2 dx
Let x=sinθ, we get
I=2∫0π/3sin2θcosθcosθ dθ
=2∫0π/3sin2θ dθ
=∫0π/3(1-cos2θ) dθ
=[θ-sin2θ2]0π/3
=π3-34
∴ 2I=2π3-32