$\int_0^\pi x \sin^7 x \cos^6 x \, dx =$

Let $H(x) = \int_{x^2}^{x^3} (x + 1) \sin(t^3) dt$. Then $\lim_{x \to 1} \frac{H(x)}{x - 1}$ is equal to:

The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

$\int_{-2}^2 (4-x^2)^{\frac{5}{2}} dx = $ (in $\pi$)

$\int_0^{\pi/6} \cos^4 3\theta \cdot \sin^2 6\theta \, d\theta$ is equal to

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