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

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

If $\int f(x) dx = F(x) + C$,then $\frac{d}{dt} \int_{g(t)}^{h(t)} f(x) dx =$

$\int_0^\pi (\sin^3 x + \cos^2 x)^2 dx = $

The value of the integral $\int_0^{\pi / 2} \sin^5 x \, dx$ is

Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ and $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$,then $(2 a+1)^{5} a^{2}$ is equal to $\dots\dots$