Integral of $\int^{2\pi}_0 \sin^{2n}(x) \cos^{2m}(x)\ dx$

Question

I have tried to solve the following integral:

$$\int^{2\pi}_0 \sin^{2n}(x) \cos^{2m}(x)\ dx$$

And obtained the following using reduction:

$$2\pi\sum_{k=0}^{m}\left(-1\right)^{m-k}\cdot\binom{m}{k}\cdot F\left(m-k+n\right)$$

where $F\left(x\right)=\frac{\left(2x\right)!}{2^{2x}\left(x!\right)^{2}}$

I have found using the double factorial: $$n!! = \prod_{i=0}^{\left\lceil\frac{n}{2}\right\rceil-1}\left(n-2i\right)$$

We can write $F(x) = \frac{\left(2x-1\right)!!}{\left(2x\right)!!}$

However, can this be simplified further? This seems like the type of integral that's been well understood for decades? What is this?

Answer 1

The elementary way by using integration by parts.

Denote $I(a,b) = \displaystyle\int_{0}^{2\pi} \sin ^{a} x\cos ^{b} x \, dx$.

Let $u=\sin ^{2n-1} x, dv = \cos ^{2m}x \sin x \, dx,$ then $du = (2n-1) \sin ^{2n-2} x\cos x \, dx, v = - \frac{\cos ^{2m+1}x}{2m+1},$ $$\begin{align*} \int_{0}^{2\pi} \sin^{2n} x\cos ^{2m}x \, dx &= -\frac{\sin ^{2n-1} x \cos ^{2m+1} x}{2m+1} \Bigg\rvert_{0}^{2\pi} + \frac{2n-1}{2m+1} \int_{0}^{2\pi} \sin ^{2n-2} x \cos ^{2m+2} x \, dx \\ &= \frac{2n-1}{2m+1} \int_{0}^{2\pi} \sin ^{2n-2} x \cos ^{2m+2} x \, dx.\end{align*}$$ So, $I(2n,2m)= \frac{2n-1}{2m+1}I(2n-2,2m+2)$ and doing integration by parts repeatedly gives $$\begin{align*} I(2n,2m)&= \frac{2n-1}{2m+1}I(2n-2,2m+2) \\ &= \frac{2n-1}{2m+1} \cdot \frac{2n-3}{2m+3}I(2n-4, 2m-4) \\ &=\cdots \\ &= \frac{(2n-1)(2n-3)\cdots (1)}{(2m+1)(2m+3)\cdots (2m+2n-1)} I(0,2n+2m).\end{align*}$$ Now, denote $Q(\alpha) = \displaystyle \int_{0}^{2\pi} \cos ^{\alpha} x \, dx$ and $I(0, 2n+2m)=Q(2n+2m)$.

Using integration by parts again, let $u= \sin x, dv= \cos ^{2m+2n-2}x \sin x \, dx,$ then $du = \cos x \, dx, v= - \frac{\cos ^{2m+2n-1}x}{2m+2n-1},$ $$\begin{align*} \int_{0}^{2\pi} \cos ^{2m+2n} x \, dx &= \int_{0}^{2\pi} \cos ^{2m+2n-2} x \, dx - \int_{0}^{2\pi} \cos ^{2m+2n-2} x \sin^{2} x \, dx \\ &= \int_{0}^{2\pi} \cos ^{2m+2n-2} x \, dx + \frac{\sin x \cos ^{2m+2n-1} x}{2m+2n-1}\Bigg\rvert_{0}^{2\pi} -\frac{1}{2m+2n-1}\int_{0}^{2\pi} \cos ^{2m+2n} x \, dx \\ &= \int_{0}^{2\pi} \cos ^{2m+2n-2} x \, dx -\frac{1}{2m+2n-1}\int_{0}^{2\pi} \cos ^{2m+2n} x \, dx.\end{align*}$$ This gives $Q(2m+2n)=\frac{2m+2n-1}{2m+2n}Q(2m+2n-2).$ Therefore, $$\begin{align*} Q(2m+2n)&=\frac{2m+2n-1}{2m+2n}Q(2m+2n-2) \\ &=\frac{2m+2n-1}{2m+2n} \cdot \frac{2m+2n-3}{2m+2n-2}Q(2m+2n-4) \\ &=\cdots \\ &=\frac{(2m+2n-1)(2m+2n-3)\cdots (1)}{(2m+2n)(2m+2n-2)\cdots (2)}\cdot 2\pi\end{align*}$$ where $Q(0)=2\pi$.

Combining altogether, we have $$\begin{align*}I(2n,2m) &= 2\pi \cdot \frac{(2n-1)!!}{2^{m+n}(m+n)!}\cdot \frac{(2m+2n-1)(2m+2n-3)\cdots (1)}{(2m+1)(2m+3)\cdots (2m+2n-1)} \\ &= \frac{\pi(2n-1)!!(2m-1)!!}{2^{m+n-1}(m+n)!}.\end{align*}$$

Answer 2

Using Gamma Function $$ \begin{aligned} \int_0^{2 \pi} \sin ^{2 n} x \cos ^{2 m} x d x = & 4 \int_0^{\frac{\pi}{2}} \sin ^{2\left(n+\frac{1}{2}\right)-1} x \cos ^{2\left(m+\frac{1}{2}\right)-1} x d x \\ = & 2 B\left(n+\frac{1}{2}, m+\frac{1}{2}\right) \\ = & \frac{2 \Gamma\left(n+\frac{1}{2}\right) \Gamma\left(m+\frac{1}{2}\right)}{\Gamma(n+m+1)} \end{aligned} $$ If $m$ and $n$ are integers, then $$ I=\frac{\pi(2 n-1) ! !(2 m-1) ! !}{2^{m+n-1}(n+m) !} $$