$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}\,d\theta = } $

If $f$ and $g$ are continuous functions on $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 2$,then $\int_0^a f(x)g(x) dx = $

The value of $\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{x^2}{1 + \tan x + \sqrt{1 + \tan^2 x}} \, dx$ is

$\int\limits_0^\infty {\frac{{{x^3}}}{{1 + x + 2{x^2} + 2{x^3} + {x^4} + {x^5}}}} dx$

The value of the integral $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x$ is

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: