3

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?

Bernard
  • 175,478
Souames
  • 293
  • 2
  • 7
  • I know I'm late but here is something, which might help: $$\int_{0}^{\frac{\pi}{2}}\sin(x)^m\cos(x)^n,\textrm{d}x=k\cdot\dfrac{(m-1)!!(n-1)!!}{(m+n)!!}$$where $k$ is $\pi/2$ itself when both $m$ and $n$ are even, and $1$ otherwise. – ultralegend5385 Aug 21 '21 at 13:35

1 Answers1

0

@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 $$I(a,b)=\frac12B\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$

clathratus
  • 17,161
  • If you're going to prove the beta function for trigonometric functions, you might as well include the proof of $$\operatorname{B}\left(m,n\right)=\frac {\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ – Frank W Dec 03 '18 at 02:50