Question

найти неопределенный интеграл
dx/$\sqrt{4-x^2$

Answer 1

$\int\limits {1/ \sqrt{4-x^2} } \, dx =1/2 \int\limits {1/ \sqrt{1-x^2/4} } \, dx = \int\limits { 1/\sqrt{1-t^2} } \, dt =$$arcsint+C=arcsin(x/2)+C$
t=x/2⇒dt=1/2dx

Answer 2

$\displaystyle \int\frac{dx}{\sqrt{4-x^2}}=\int\frac{dx}{\sqrt{4\left(1-\frac{x^2}{4}\right)}}=\int\frac{dx}{2\sqrt{1-\frac{x^2}{4}}}=\int\frac{dx}{2\sqrt{1-\left(\frac{x}{2}\right)^2}}=$
$\displaystyle =\int\frac{\frac{1}{2}dx}{\sqrt{1-\left(\frac{x}{2}\right)^2}}=\int\frac{d\left(\frac{x}{2}\right)}{\sqrt{1-\left(\frac{x}{2}\right)^2}}=arcsin\left(\frac{x}{2}\right)+C.$