If $I = \int_0^{\pi /4} \sin^2 x \, dx$ and $J = \int_0^{\pi /4} \cos^2 x \, dx$,then $I = $

$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt[3]{\sec x}}{\sqrt[3]{\sec x}+\sqrt[3]{\operatorname{cosec} x}} d x=$

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} {x\cot x\,dx} $ equals

$\left|\frac{120}{\pi^3} \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx\right|$ is equal to:

The value of the definite integral $\int_{\pi / 24}^{5 \pi / 24} \frac{d x}{1+\sqrt[3]{\tan 2 x}}$ is