Question

Вычислить определенный интеграл

Answer 1

$\displaystyle \int\limits^\frac{1}{\sqrt2}_0\frac{arccos^3x-1}{\sqrt{1-x^2}}dx=-\int\limits^\frac{1}{\sqrt2}_0(arccos^3x-1)d(arccosx)=\\=-\frac{arccos^4x}{4}+arccosx|^\frac{1}{\sqrt2}_0=-\frac{\pi^4}{4^5}+\frac{\pi}{4}+\frac{\pi^4}{64}-\frac{\pi}{2}=\frac{15\pi^4}{1024}-\frac{\pi}{4}\approx0,641$