$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx =$

$\int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx =$

Let $f(x)$ be integrable over $(a, b)$,where $b > a > 0$. If $I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \sec^2 \theta \, d\theta$ and $I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \csc^2 \theta \, d\theta$,then the ratio $\frac{I_1}{I_2}$ is:

For $x > 0,$ if $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt,$ then $f(e) + f\left(\frac{1}{e}\right)$ is equal to:

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 $R_2$ is:

$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$