$I=\displaystyle\int \sqrt{3-x^2} \,dx\\ x=\sqrt{3}\sin t \Rightarrow t=\arcsin \dfrac{x}{\sqrt{3}}\\ dx=\sqrt{3}\cos t dt\\ I=\displaystyle\int \sqrt{3-3\sin^2t} \,\sqrt{3}\cos t dt\\ =\displaystyle\int 3\cos^2t dt\\ =3\displaystyle\int \dfrac{1+\cos(2t)}{2} dt\\ =\dfrac{3}{2}\displaystyle\int dt +\dfrac{3}{4}\displaystyle\int\cos(2t) d(2t)\\ =\dfrac{3}{2}t+\dfrac{3}{4}\sin(2t)+C\\ =\dfrac{3}{2}\arcsin \dfrac{x}{\sqrt{3}}+\dfrac{3}{4}\sin\left(2\arcsin \dfrac{x}{\sqrt{3}}\right)+C$
