Question

помогите решить интеграл arccos2xdx

Answer 1

$\int arccos2xdx=x \cdot arccos2x-\int xd(arccos2x)= \\ =x \cdot arccos2x-\int xd(arccos2x)= x \cdot arccos2x+\int \frac{2xdx}{\sqrt{1-4x^2}} = \\ = x \cdot arccos2x- \frac{1}{4} \int \frac{d(1-4x^2)}{\sqrt{1-4x^2}} = x \cdot arccos2x- \\-\frac{1}{4} \int (1-4x^2)^{-0,5} d(1-4x^2)= x \cdot arccos2x- \frac{1}{2} \sqrt{1-4x^2}+C$