Let $I_n = \int_{0}^{\frac{\pi}{4}} \tan^n x \, dx$. Then $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \dots$ are in:

Let $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$ (where $\alpha > 0$),then $\sum_{\alpha=1}^{5} \frac{f(\alpha)}{\alpha^{3}}$ is equal to-

Let $[t]$ denote the greatest integer less than or equal to $t.$ Then,the value of the integral $\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$ is equal to

The value of $\int\limits_0^{\frac{\pi }{2}} {\sin 8x \cot x \, dx} + \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)dx}$ is equal to

The numbers $P, Q$ and $R$ for which the function $f(x) = P{e^{2x}} + Q{e^x} + Rx$ satisfies the conditions $f(0) = -1$,$f'(\log 2) = 31$ and $\int_0^{\log 4} [f(x) - Rx] \, dx = \frac{39}{2}$ are given by

$\int_{0}^{1/3} (\sum_{r=0}^{101} \{x + \frac{r}{3}\}) dx$ is equal to (where $\{.\}$ represents the fractional part function).