Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

If $\alpha = 1$ and $\beta = 1 + i\sqrt{2}$, where $i = \sqrt{-1}$ are two roots of the equation $x^3 + ax^2 + bx + c = 0$ with $a, b, c \in R$, then $\int_{-1}^{1} (x^3 + ax^2 + bx + c) dx$ is equal to:

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

$\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x=$

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$,$x \in R - \{-1\}$,$n \in N$,$n > 2$. If $f^n(x) = (f \circ f \circ f \dots \text{upto } n \text{ times})(x)$,then $\lim_{n \to \infty} \int_0^1 x^{n-2} (f^n(x)) dx$ is equal to $...............$.

The value of the integral $\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$,where $p$ and $q$ are integers,is equal to: