Problem

$y = \arccos(2x)$

(a) Determine

$$\frac{dy}{dx}.$$

(b) Show that

$$\int y\,dx = x\arccos(2x) + \int \frac{x}{\sqrt{1 - 4x^2}}\,dx.$$

(c) Use the substitution $x = \frac{1}{2}\cos\theta$ to show that

$$\int \frac{x}{\sqrt{1 - 4x^2}}\,dx = \frac{1}{4}\int_a^b \cos 2\theta\,d\theta,$$

where $a$ and $b$ are limits to be determined.

(d) Hence, determine the exact value of

$$\int_0^{1/8} \arccos(2x)\,dx.$$