If $\int f(x) dx = F(x) + C$,then $\frac{d}{dt} \int_{g(t)}^{h(t)} f(x) dx =$

Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in \left(0, \frac{\pi}{2}\right)$,if $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$,then $f\left(\frac{1}{\sqrt{2}}\right)=$

$\mathop {Limit}\limits_{x \to {x_1}} \,\,\frac{x}{{x - {x_1}}}\,\,\int\limits_{{x_1}}^x {f(t)} \, dt$ 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

$\int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$ equals

$\int_0^1 x^{3/2} \sqrt{1-x} \, dx$ is equal to