If $\int_{\sin x}^1 {{t^2}f(t)\;dt = 1 - \sin x} $,$x \in \left( {0,\frac{\pi }{2}} \right)$,then $f\left( {\frac{1}{{\sqrt 3 }}} \right)$ is equal to:

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$,if $F'(a) + 2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$,then $f(0)$ is

The number of solutions of the equation $\frac{d}{dx} \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}} = 2\sqrt{2}$ in the interval $[0, \pi]$ is:

If $I_n = \int_0^a \frac{x^n}{\sqrt{a^2-x^2}} dx$, then $\frac{I_8}{I_4} =$

$\int_0^{\pi / 2} \sin ^8 x \cos ^2 x \, dx$ is equal to

If $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$,then