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 $f\left(\frac{1}{\sqrt{2}}\right)=$

$\int_0^{\pi /2} \sin^5 x \, dx = $

$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ is equal to

If $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0,$ then $\frac{dy}{dx} = $

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 points of extremum of $\int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$ are