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

$\lim \limits_{x \rightarrow 1} \left( \frac{\int \limits_{0}^{(x-1)^{2}} t \cos(t^{2}) dt}{(x-1) \sin(x-1)} \right)$ is equal to

The value of $x$ that maximises the value of the integral $\int\limits_x^{x + 3} {t(5 - t)\,dt}$ is

Let $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$,then $\int_0^{\pi /2} f(x) dx = $

$\int\limits_0^{\frac{\pi }{4}} (\cos 2x)^{3/2} \cos x \,dx =$

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are