If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then

Let $f(x)$ be integrable over $(a, b)$,where $b > a > 0$. If $I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \sec^2 \theta \, d\theta$ and $I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \csc^2 \theta \, d\theta$,then the ratio $\frac{I_1}{I_2}$ is:

$\int_{-1}^{1} x^{17} \cos^{4} x \, dx = $

The function $F(x) = \int_0^x \log \left( \frac{1 - t}{1 + t} \right) \,dt$ is

If $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,then
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$

Let $f$ and $g$ be continuous functions on $[0, a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$,then $\int_0^a f(x) g(x) d x$ is equal to