I've recently discovered Wallis' formula to compute powers of cos and sine from $[0,\pi/2]$, However what If I have a function like $\cos^m (x)\sin^n(x)$ where both $m$ and $n$ are even, this function is even, so it must be symmetric to some axis.
If I want to compute the integral of this function but from $[0,k\pi]$. Can I use parity of the function to integrate from $0$ to $\pi/2$ and then use Wallis formula?
Also I've noticed that for $\cos^m (x)\sin^n(x)$ if one of $m$ or $n$ is odd then the integral on $[0,k\pi]$ is $0$, why is that?
@Zacky showed you that identity with the Beta function. Here's how to derive it.
Consider the integral $$I(a,b)=\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx$$ Straight away, we preform the substitution $t=\sin(x)^2$: $$ \begin{align} I(a,b)=&\frac12\int_0^1t^{a/2}(1-t)^{b/2}t^{-1/2}(1-t)^{-1/2}\mathrm dt\\ =&\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt\\ =&\frac12\int_0^1t^{\frac{a+1}2-1}(1-t)^{\frac{b+1}2-1}\mathrm dt\\ \end{align} $$ Next we recall the definition of the Beta function: $$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Hence we have $$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$