Verified answer

Firstly, in this case, the square brackets are used to distinguish from round brackets.

Now, with the integral above, use integration by parts using u=x^(2n) and dv = sin(pi x) dx.

You get : -(1/pi) x^(2n) cos(pi x) + (2n/pi) Integral[x^(2n-1) cos(pi x) dx]

For this other integral, use int. by parts again, with u=x^(2n-1) and dv=cos(pi x) dx. You get :

(1/pi) x^(2n-1) sin(pi x) - ((2n-1)/pi) Integral[x^(2n-2) sin(pi x) dx]

Putting all this together:

-(1/pi) x^(2n) cos(pi x) + (2n/pi) { (1/pi) x^(2n-1) sin(pi x) - ((2n-1)/pi) Integral[x^(2n-2) sin(pi x) dx] }

Evaluating between x=0 and x=1, the term with sin(pi x) (outside the integrals) will vanish, and cos(pi x) will be 1 and -1 respectively. This will give :

J(n) = -(1/pi)[(1)^(2n) (-1) - 0] - [(2n)(2n-1)/pi^2] J(n-1)

J(n) = (1/pi) - [(2n)(2n-1)/pi^2] J(n-1) = (1/pi^2) [ pi - (2n)(2n-1) J(n) ]