$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \left(\frac{\pi-x}{\pi+x}\right) \cdot \cos x\right) dx =$

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

If $A=\int_0^{\infty} \frac{1+x^2}{1+x^4} d x$ and $B=\int_0^1 \frac{1+x^2}{1+x^4} d x$,then

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$,and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $x$-axis. Then

If $[.]$ denotes the greatest integer function,then $\int_0^{1000} e^{x-[x]} dx=$

If ${I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx}$,then $\mathop {\lim }\limits_{n \to \infty } \,n({I_n} + {I_{n - 2}})$ equals