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

Given that $\frac{d}{d x}\left[\int_0^{\phi(x)} f(t) d t\right]=f(\phi(x)) \cdot \phi^{\prime}(x)$. If $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$,then the value of $f(8)$ is

$\int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x \, dx =$

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 value of $\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to:

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