The value of the integral $\int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$ is equal to

Evaluate the definite integral $\int_{-1}^{2} \left[ \frac{[x]}{1 + x^2} \right] dx$,where $[\cdot]$ denotes the Greatest Integer Function $(GIF)$.

Choose the correct option regarding the following definite integrals:
$(i)$ $\int_0^{\pi / 2} \sin ^m(x) \cos (x) d x = \frac{1}{m+1}$
(ii) $\int_0^{\pi / 2} \sin (x) \cos ^n(x) d x = \frac{1}{n+1}$

If the value of the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,dx = \alpha e^{-1} + \beta$,where $\alpha, \beta \in R, 5\alpha + 6\beta = 0$,and $[x]$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha + \beta)^{2}$ is equal to:

The value of the definite integral $\int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $ is

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$, using the Trapezoidal rule, is:

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$