Let y = y(x) be the solution of the differential equation

$$\sin x{{dy} \over {dx}} + y\cos x = 4x$$, $$x \in \left( {0,\pi } \right)$$.

If $$y\left( {{\pi \over 2}} \right) = 0$$, then $$y\left( {{\pi \over 6}} \right)$$ is equal to :

The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$$ is

The integral

$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$$

is equal to

Let g(x) = cosx², f(x) = $$\sqrt x $$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x² - 9$$\pi $$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve
y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is :