Question

Write a value of $\int \sqrt{4-x^2}dx$.

Answer 1

$$\begin{gathered} \int \sqrt{4-x^2}dx \\ =\int \sqrt{2^2-x^2}dx \\ =\frac{x}{2}\sqrt{2^2-x^2}+\frac{2^2}{2}{\sin }^{-1}\left(\frac{x}{2}\right)+C\left(\because \sqrt{a^2-x^2}=\frac{x}{2}\sqrt{a^2-x^2}-\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C\right) \\ =\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\left(\frac{x}{2}\right)+C \end{gathered}$$