The value of the integral $$\int\limits_{ - \pi /2}^{\pi /2} {\left( {{x^2} + 1n{{\pi + x} \over {\pi - x}}} \right)\cos xdx} $$ is

The value of $$\,\int\limits_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {{{x\sin {x^2}} \over {\sin {x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}\,dx} $$ is

Let the straight line $$x=b$$ divide the area enclosed by
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals

Let f $$:$$$$\left[ { - 1,2} \right] \to \left[ {0,\infty } \right]$$ be a continuous function such that
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$

Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,} $$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then