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

$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ is equal to

The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) dt - (x^2 - x + 1) e^x, x \in R$,is.

If $F(x) = \int_{x^2}^{x^3} \log t \, dt$ $(x > 0)$,then $F'(x) = $

The value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$ is

Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in \left(0, \frac{\pi}{2}\right)$,if $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$,then $f\left(\frac{1}{\sqrt{2}}\right)=$