The value of the integral $$\int\limits_0^1 {\sqrt {{{1 - x} \over {1 + x}}} dx} $$ is

The area enclosed between the curves $$y = a{x^2}$$ and
$$x = a{y^2}\left( {a > 0} \right)$$ is $$1$$ sq. unit, then the value of $$a$$ is

If $$f(x)$$ is differentiable and $$\int\limits_0^{{t^2}} {xf\left( x \right)dx = {2 \over 5}{t^5},} $$ then $$f\left( {{4 \over {25}}} \right)$$ equals

If $$y=y(x)$$ and $${{2 + \sin x} \over {y + 1}}\left( {{{dy} \over {dx}}} \right) = - \cos x,y\left( 0 \right) = 1,$$
then $$y\left( {{\pi \over 2}} \right)$$ equals