The maximum distance from origin of a point on the curve
$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is

$${I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} $$ then $$\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]$$ equals

$$\int\limits_0^2 {\left[ {{x^2}} \right]dx} $$ is

If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx} $$ is