The integral
$$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $$
is equal to:

Let $$f(x) = \left| {x - 2} \right|$$ and g(x) = f(f(x)), $$x \in \left[ {0,4} \right]$$. Then
$$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$$ is equal to:

If the value of the integral
$$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$$

is $${k \over 6}$$, then k is equal to :

Suppose f(x) is a polynomial of degree four, having critical points at –1, 0, 1. If
T = {x $$ \in $$ R | f(x) = f(0)}, then the sum of squares of all the elements of T is :