$\int_{0}^{1} x(1-x)^{5} dx =$

The value of $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$,where $[t]$ denotes the greatest integer less than or equal to $t$,is

Let $a$ be a positive real number such that $\int_{0}^{a} e^{x-[x]} dx = 10e - 9$,where $[x]$ is the greatest integer less than or equal to $x$. Then $a$ is equal to:

The value of the integral $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

$\int\limits_0^\pi {{e^{{{\cos }^4}x}}} \cdot \cos^5(2n + 1)x \,dx, (n \in I)$ is equal to