General case integral $\int^1_0 f(x) \, f'(1-x) \, dx $ Suppose we get the following general case 
$$\int^1_0 \, f(x) \, f'(1-x) \, dx $$
where $f$ is differentiable on $[0,1]$ . Can we have a general formula for the integral ? 
Of cousre we can generalize for
$$\int^a_0 \, f(x) \, f'(a-x) \, dx $$ 
Interesting examples would be 
$$\int^{\frac{\pi}{2}}_0 \sin^n(x) dx $$
Which is easily found by beta function or integration be parts .
 A: Here is a suggestion for functions of class $\mathcal{C}^1$:
Suppose first that $f(0)=f(1)$. In that case, $f$ can be extended into a 1-periodic function via a Fourier series $$f(x)=\sum_{-\infty}^\infty c_n e^{2 \pi n i x}, \\ c_n=\int_0^1f(x) e^{-2 \pi n ix }\mathrm{d} x.$$
The derivative is then $$f'(x)= \sum_{-\infty}^\infty 2 \pi n i c_n e^{2 \pi n i x},$$ so that $$f'(1-x)=\sum_{-\infty}^\infty2 \pi n i c_ne^{-2 \pi n i x}. $$
When multiplying $f(x) \cdot f'(1-x)$ and integrating over the period, only the constant coefficient is of interest. The integral is then $$ \int_0^1 f(x) f'(1-x) \mathrm{d} x= \sum_{-\infty}^\infty 2\pi n i c_n^2 .$$
If $f(0) \neq f(1)$, consider the function $g(x)=f(x)-x(f(1)-f(0))$. It is of class $\mathcal{C}^1$ as well as $f$, and it also satisfies $g(0)=g(1)$. $g$ has Fourier coefficients $$d_n=c_n-(f(1)-f(0))e_n ,$$ where $$e_n=\begin{cases} \frac{1}{2}& n=0 \\ \frac{i}{2 \pi n} & n \neq 0 \end{cases} .$$
We know that $$\int_0^1 g(x) g'(1-x) \mathrm{d} x= \sum_{-\infty}^\infty 2 \pi n i d_n^2 .$$
The integral can be further simplified: $$\int_0^1 [f(x)-x(f(1)-f(0))][f'(1-x)-(f(1)-f(0))] \mathrm{d} x \\=\int_0^1 f(x) f'(1-x) \mathrm{d} x-(f(1)-f(0)) \int_0^1 f(x) \mathrm{d} x-(f(1)-f(0)) \int_0^1 x f'(1-x) \mathrm{d} x+\frac{(f(1)-f(0))^2}{2}. $$ We recognize $\int_0^1 f(x) \mathrm{d} x$ as $c_0$, and integrate $\int_0^1 x f'(1-x) \mathrm{d} x$ by parts. Finally we have $$\int_0^1 f(x) f'(1-x) \mathrm{d} x=(f(1)-f(0))(2c_0-f(0))-\frac{(f(1)-f(0))^2}{2}+ \sum_{-\infty}^\infty 2 \pi n i d_n^2. $$
