If $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$ for all $x \in R - \{0\}$,then the value of $f(\frac{\pi}{6})$ is

Let $f : (-1, 1) \to R$ be a continuous function. If $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$,then $f\left(\frac{\sqrt{3}}{2}\right)$ is equal to

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

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

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \right)$ is