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 the function $f: R \rightarrow R$ be defined by
$f(t)=\begin{cases} (-1)^{n+1} 2, & \text{if } t=2n-1, n \in N \\ \frac{(2n+1-t)}{2} f(2n-1) + \frac{(t-(2n-1))}{2} f(2n+1), & \text{if } 2n-1 < t < 2n+1, n \in N \end{cases}$
Define $g(x) = \int_1^x f(t) dt, x \in (1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x) = 0$ in the interval $(1, 8]$ and $\beta = \lim_{x \rightarrow 1^+} \frac{g(x)}{x-1}$. Then the value of $\alpha + \beta$ is equal to.

Let $a, b, c$ be non-zero real numbers such that $\int_0^1 {(1 + \cos^8 x)(ax^2 + bx + c) \, dx} = \int_0^2 {(1 + \cos^8 x)(ax^2 + bx + c) \, dx}$. Then the quadratic equation $ax^2 + bx + c = 0$ has:

Let $f: R \rightarrow R$ be defined as $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1$,$f^{\prime}(\log _e 2)=21$,and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$,then the value of $|a+b+c|$ equals:

If $m \in Z^{+}$,$n=2m$ and $\int_0^{\frac{\pi}{2}} \sin ^{m} x \cos ^{n} x \, dx = K(m) \int_0^{\frac{\pi}{2}} \sin ^m x \, dx$,then $\frac{2^{m-1}(m-1)!}{(2m-1)!} K(m) =$

If $A_n = \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^n x \, dx$,then $\frac{A_4 - A_6}{A_4} = $