Let Jn= the integral between 0 and 1 of (x^2n)(sin(Pi*x)) .dx where n=1,2,3....

Use integration by parts to show that:

Jn+1 = 1/Pi^2 [Pi - (2n+1)(2n+2) *Jn]

So I started by setting u=sin(Pi*x) and dv=x^(2n+2)

so that du=Pi*cos(Pi*x) and v= (x^(2n+1))/(2n+1)

But this just seems to make the integral harder when I put it together, and so does doing it the other way round. What am I doing wrong? Please could I have some help with this.