Question:
$\sqrt{4-x^{2}}$
Solution:
Let $I=\int \sqrt{4-x^{2}} d x=\int \sqrt{(2)^{2}-(x)^{2}} d x$
It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+\mathrm{C}$
$\begin{aligned} \therefore I &=\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}+\mathrm{C} \\ &=\frac{x}{2} \sqrt{4-x^{2}}+2 \sin ^{-1} \frac{x}{2}+\mathrm{C} \end{aligned}$
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