The values of $\alpha$ which satisfy $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$,where $\alpha \in [0, 2\pi]$,are equal to:

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

If ${I_n} = \int_{ - n}^n {{{\tan }^2}\{x\}dx} $ then (where $\{.\}$ denotes the fractional part function and $n \in N$):

If the abscissa of the vertex of the parabola $y = ax^2 + bx + c$ is $1$ $(a, b, c > 0)$ and $f(x) = \int_0^x (3at^2 + bt + c) dt$ is a strictly increasing function $\forall x \in R$,then the maximum possible value of $[\frac{a}{c}]$ is (where $[.]$ denotes the greatest integer function).

If $f(x) = \text{Max}\{\sin x, \cos x\}$ and $g(x) = \text{Min}\{\sin x, \cos x\}$,then $\int_{0}^{\pi} f(x) dx + \int_{0}^{\pi} g(x) dx = $

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be a function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_{0}^{1} f(x)g(x) \, dx$ is