The value of the integral $\int_{0}^{0.9} [x - 2[x]] \, dx$,where $[.]$ denotes the greatest integer function,is

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

$\int_0^1 \frac{x}{(1-x)^{3/4}} dx = $

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

$\int_0^{\pi /2} {\frac{{\sin x\cos x\,dx}}{{{{\cos }^2}x + 3\cos x + 2}}} = $

If $I=\int_0^{\pi / 2} \frac{d x}{5+3 \sin x}=\lambda \tan ^{-1}\left(\frac{1}{2}\right)$, then $\lambda=$