$\int_0^1 x(1-x)^n dx =$

If $f(a + b - x) = f(x)$,then $\int_{a}^{b} x \cdot f(a + b - x) \, dx = $

Let $f : R \rightarrow R$ be a continuous function satisfying $f(x) + f(x + k) = n$,for all $x \in R$,where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4nk} f(x) dx$ and $I_{2} = \int_{-k}^{3k} f(x) dx$,then:

If $I = \int_0^{\frac{\pi}{2}} \cos(\sin x) \,dx$,$J = \int_0^{\frac{\pi}{2}} \sin(\cos x) \,dx$,and $K = \int_0^{\frac{\pi}{2}} \cos x \,dx$,then:

The value of the integral $\int_0^\pi(1-|\sin 8 x|) d x$ is

$\int_0^{\pi / 2} |\sin t - \cos t| \, dt =$