If $A_n = \int_{0}^{\pi /2} \frac{\sin((2n-1)x)}{\sin x} dx$ and $B_n = \int_{0}^{\pi /2} \left( \frac{\sin(nx)}{\sin x} \right)^2 dx$ for $n \in N$,then:

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$,and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$,then the value of $f(0)$ is.

If $f(x) = \text{Max}\{\sin x, \cos x\}$ and $g(x) = \text{Min}\{\sin x, \cos x\}$,then $\int_{0}^{\pi} f(x) dx + \int_{0}^{\pi} g(x) dx = $

Let $I_n = \int_0^1 e^{-y} y^n \, dy$,where $n$ is a non-negative integer. Then,$\sum_{n=1}^{\infty} \frac{I_n}{n!}$ is

$ 6\int_{0}^{\pi}|(\sin 3x+\sin 2x+\sin x)| dx $ is equal to ....

The value of the limit $\lim _{n \rightarrow \infty} \int _{0}^{1} x^{10} \sin (n x) d x$ equals