$${I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} $$ 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

$$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$$ 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