2.5 посмотрите пожалуйста

Решите задачу:

$\int\limits_{-\frac12}^{\frac12}\arccos2xdx=\left(\begin{array}{c}u=2x\\du=2dx\end{array}\right)=\frac12\int\limits_{-\frac12}^{\frac12}\arccos udu=\\=\frac12\left.\left(u\arccos u-\sqrt{1-u^2}\right)\right|\limits_{-\frac12}^{\frac12}=\left(\begin{array}{c}u=2x\end{array}\right)=\\=\left.\frac12(2x\arccos2x-\sqrt{1-(2x)^2})\right|\limits_{-\frac12}^{\frac12}=\left.(x\arccos2x-\frac12\sqrt{1-4x^2})\right|\limits_{-\frac12}^{\frac12}=$
$=(\frac12\arccos1-\frac12\sqrt0)-(-\frac12\arccos(-1)-\frac12\sqrt0)=\\=(0-0)-(-\frac12\cdot\pi-0)=\frac\pi2$