$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x$ is equal to

Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

If $I_{n+1} = \int_{0}^{1} \frac{x^{n+1} - 1}{x + 1} dx$,then the value of $I_{10} + I_{11} + 2 \log 2$ is:

If $\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx = \frac{1}{a} \log_e\left(\frac{a}{3}\right) + \frac{\pi}{b \sqrt{3}}$,where $a, b \in N$,then $a+b$ is equal to ....................

$\int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} = $

$ \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $ is equal to