If $f(x) = \int_0^x {t(\sin x - \sin t) dt}$,then which of the following is true?

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be a function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_{0}^{1} f(x)g(x) \, dx$ is

If $A_n = \int_{\frac{\pi}{2}}^{\infty} e^{-x} \cos^n x \, dx$,then $\frac{A_4 - A_6}{A_4} = $

If $b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2} nx}{\sin x} dx$,$n \in N$,then

The tangent to the graph of the function $y = f(x)$ at the point with abscissa $x = a$ forms with the $x$-axis an angle of $\pi/3$ and at the point with abscissa $x = b$ at an angle of $\pi/4$. Then the value of the integral $\int_{a}^{b} f(x) \cdot f''(x) \, dx$ is equal to (assume $f''(x)$ to be continuous).

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$ is