Assuming [tex]n\in\mathbb Z[/tex]:
[tex]\displaystyle\int_0^\pi\cos^{2n+1}x\,\mathrm dx=\int_0^\pi\cos^{2n}x\cos x\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^\pi(1-\sin^2x)^n\cos x\,\mathrm dx[/tex]
Let [tex]y=\sin x[/tex], so that [tex]\mathrm dy=\cos x\,\mathrm dx[/tex]. The integral is then
[tex]=\displaystyle\int_{\sin0=0}^{\sin\pi=0}(1-y^2)^n\,\mathrm dy=0[/tex]
