$\int_{-1}^{1} x|x| \, dx = $

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$,then $\frac{1}{I_2 + I_4} + \frac{1}{I_3 + I_5} + \frac{1}{I_4 + I_6} = $

Let $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ and $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,then:

By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

If $\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin ^4 x)} = \alpha \pi + \beta \log _e(3+2 \sqrt{2})$,where $\alpha, \beta$ are integers,then $\alpha^2+\beta^2$ equals.....................

$\int_{-1}^{1} \sin^{11} x \, dx$ is equal to