$$\int_\limits 0^\pi \frac{x \tan x}{\sec x+\cos x} d x= $$

$$\int_\limits0^1 \cos ^{-1} x d x=$$

If $$\int_\limits0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6}$$, then the value of $$\mathrm{k}$$ is

$$\int_\limits1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5} \text {, then } \mathrm{k} \text { has the value }$$