For instance, let $m$ be an odd number. Then the Fourier series of $\sin^m(x) \cos^n(x)$ have forms like $\displaystyle \sum_{k=1}^{+\infty}b_k \sin(kx)$. Next, we can get $b_k$ through the integral $$\displaystyle b_k = \frac{1}{\pi}\int_{-\pi}^{\pi} \sin^m(x)\cos^n(x)\sin(kx) \mathrm{d}x$$ However, I find it too hard to compute. Even though I use formulas like $$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^p(x)\cos^q(x) \mathrm{d} x=\dfrac{\Gamma\left(\frac{p+1}{2}\right)\Gamma\left(\frac{q+1}{2}\right)}{2\Gamma\left(\frac{p+q}{2}+1\right)}$$ and $$\displaystyle\sin(kx)=\sum_{j=0}^{[\frac{k-1}{2}]}(-1)^j\binom{k}{2j+1}\cos^{k-2j-1}(x)\sin^{2j+1}(x)$$ I still get a bulk of sums and factorials: $$\displaystyle b_k = \dfrac{1}{2^{\frac{k+n+m}{2}-3}}\sum_{j=0}^{[\frac{k-1}{2}]}(-1)^j\dfrac{k!(m+2j)!!(k+n-2j-2)!!}{(2j+1)!(k-2j-1)!\left(\frac{k+n+m}{2}\right)!}, \text{ when } n \text{ and } k \text{ are not both odd or even}$$ $$b_k = 0, \text{ when } n \text{ and } k \text{ are both odd or even}$$ What's worse, I can't even get $b_k = 0$ when $k>m+n$ through this way, which should theoretically hold. I am not familiar with skills of calculating factorials so I can only use computer programs to check $b_k = 0$ when $k>m+n$.
Is it possible to work out the Fourier series of $\sin^m(x) \cos^n(x)$ without knowing the positive integers $m$ and $n$?
If you know a bit of complex analysis you can use the residue theorem for calculating trigonometric integrals. I will use the bases of complex exponentials $$ I=\int_{-\pi}^\pi\sin^m(x)\cos^n(x) e^{ikx} dx=\frac{1}{2^{m+n} i^{m+1}}\int_C \frac{(z^2-1)^m (z^2+1)^n }{z^{m+n+1-k}} dz $$ Notice that if $k\ge m+n+1$ it is the integral of an holomorphic function and so it is zero
For all the other values of $k$ the integral is evaluated as $$ \frac{\pi}{2^{m+n-1}i^m(m+n-k)!} \lim_{z \to 0} \frac{d^{m+n-k}}{dz} (z^2-1)^m (z^2+1)^n $$ As the degree of the polynomial is $2(m+n)$ if $k\le -(m+n+1)$ the residue is zero.
Moreover as the polynomial has only terms of even degree, if $m+n-k$ is odd the residue is zero (we are taking the limit for $z\to0$).
You have
$$ (z^2-1)^m (z^2+1)^n =\left(\sum_{i=0}^m (-1)^{m-i}\binom{m}{i} z^{2i}\right) \left(\sum_{j=0}^n \binom{n}{j} z^{2j}\right) $$ By using the (finite) Cauchy product, calculating the derivative and the limit we obtain $$ I=\frac{\pi}{2^{m+n-1} i^m} \sum_{\substack{l_1+l_2=\frac{m+n-k}{2}\\l_1 \le m, \ \ l_2 \le n}} (-1)^{m-l_1} \binom{m}{l_1} \binom{n}{l_2} $$ for $-(m+n+1) < k < m+n+1$ such that $k$ has the same parity w.r.t. $m+n$
If you want the expansion to be in the bases of $(\sin(kx),\cos(kx))$ you can then proceed as here