Let $x$ be the length of each of the equal sides of an isosceles triangle and $\theta$ be the angle between these sides. If $x$ is increasing at the rate $\frac{1}{12} \mathrm{~m} /$ hour and $\theta$ is increasing at the rate $\frac{\pi}{180} \mathrm{rad} /$ hour, then the rate at which area of the triangle is increasing when $x=12 \mathrm{~m}$ and $\theta=\frac{\pi}{4}$ is

$$ \int \cos \left(\frac{x}{16}\right) \cdot \cos \left(\frac{x}{8}\right) \cdot \cos \left(\frac{x}{4}\right) \cdot \sin \left(\frac{x}{16}\right) \mathrm{d} x= $$

$$ \int \frac{x^3}{(x+1)^2} d x= $$

A wire of length 8 units is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is