If $f(x) = f(\Pi + e - x)$ and $\int_{e}^{\Pi} f(x) dx = \frac{2}{e + \Pi}$,then $\int_{e}^{\Pi} x f(x) dx$ is equal to

Evaluate the definite integral: $\int_0^{\pi /2} \log(\tan x) \, dx$.

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$ for $\alpha > 0$ is

Let $g(t) = \int_{-\pi/2}^{\pi/2} \cos \left(\frac{\pi}{4} t + f(x)\right) \, dx$,where $f(x) = \log_e \left(x + \sqrt{x^2 + 1}\right)$,$x \in R$. Then which one of the following is correct?

$\left[ {\sum\limits_{n = 1}^{10} {\int_{ - 2n - 1}^{2n} {{{\sin }^{27}}x\,dx} } } \right] + \left[ {\sum\limits_{n = 1}^{10} {\int_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} } } \right]$ equals

$\int_0^{2 \pi} (\sin x + |\sin x|) \, dx =$