If $S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} dx$ and $n$ is an integer,then $S_{n+1} - S_n =$

If $\int_0^{\pi / 2} \tan ^{14}\left(\frac{x}{2}\right) d x=2\left[\sum_{n=1}^7 f(n)-\frac{\pi}{4}\right]$, then $f(n)=$

$\int_{0}^{1} (1 + |\sin x|)(ax^2 + bx + c) dx = \int_{0}^{2} (1 + |\sin x|)(ax^2 + bx + c) dx$. Then,the location of the roots of $ax^2 + bx + c = 0$ is:

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$,and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $x$-axis. Then

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\sin^{2} x}{1+\pi^{\sin x}}\right) \, dx$ is

$\int_0^{\pi / 2} \frac{16 x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$ is equal to