Question

Integrate the function.
$\int \sqrt{4-x^2}dx.$

Answer 1

Let $I=\int \sqrt{4-x^2}dx=\int \sqrt{2^2-x^2}dx$
$=\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}si{n}^{-1}\frac{x}{2}+C[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}+C]\Rightarrow I=\frac{x}{2}\sqrt{4-x^2}+2si{n}^{-1}\frac{x}{2}+C$