If f : R $$ \to $$ R is a differentiable function and f(2) = 6,
then $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$ is :-

The value of the integral $$\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $$ is :-

The value of $$\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} $$ is

Let $$f(x) = \int\limits_0^x {g(t)dt} $$ where g is a non-zero even function. If ƒ(x + 5) = g(x), then $$ \int\limits_0^x {f(t)dt} $$ equals-