The angle of intersection between the curves $y = \int\limits_{x^2}^{x^3} \sqrt{5 - t^2} \, dt$ and the $x$-axis is (where $x \neq 0$):

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

Let $f$ be a continuous function satisfying $\int \limits_0^{t^2} (f(x) + x^2) dx = \frac{4}{3} t^3, \forall t > 0$. Then $f \left(\frac{\pi^2}{4}\right)$ is equal to:

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 $f$ be a differentiable function in $\left(0, \frac{\pi}{2}\right)$. If $\int\limits_{\cos x}^{1} t^{2} f(t) d t = \sin^{3} x + \cos x - 1$,then $\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f(\frac{tx}{36}) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points $(2, 1)$ and $(-4, \beta)$, then $\beta^{\alpha}$ is equal to . . . . . . .