$\int \sqrt{4-9 x^2}\mathit{\ⅆ}x$

$\= \int 2 \cos \left(\mathrm{\θ}\right)\cdot \left(\frac{2}{3}\cos \left(\mathrm{\θ}\right)d\mathrm{\θ}\right)$

$\= \frac{4}{3}\int {\cos }^2\left(\mathrm{\θ}\right) d\mathrm{\θ}$

$\=\frac{4}{3}\int \frac{1}{2}\left(1\+\cos \left(2 \mathrm{\θ}\right)\right) \mathit{\ⅆ}\mathrm{\θ}$

$\=\frac{2}{3}\left(\mathrm{\θ}\+\frac{1}{2}\sin \left(2 \mathrm{\θ}\right)\right)$

$\=\frac{2}{3}\left(\mathrm{\θ}\+\sin \left(\mathrm{\θ}\right)\cos \left(\mathrm{\θ}\right)\right)$

$\=\frac{2}{3}\left(\arcsin \left(\frac{3}{2}x\right)\+\left(\frac{3}{2}x\right)\left(\frac{1}{2}\sqrt{4-9 x^2}\right)\right)$

$\=\frac{x}{2}\sqrt{4-9 x^2}\+\frac{2}{3}\arcsin \left(\frac{3}{2}x\right)$

Evaluate the given integral

$\int \sqrt{4-9 x^2}\mathit{\ⅆ}x$ = $\frac{1}{2}x\sqrt{-9x^2\+4}\+\frac{2}{3}\arcsin \left(\frac{3}{2}x\right)$$$

Initialization

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$

$Q≔\int \sqrt{4-9 x^2}\mathit{\ⅆ}x\:$

Change variables as per Table 6.3.1

$q_1≔\mathrm{Change}\left(Q\,x\=\frac{2}{3}\sin \left(\mathrm{\θ}\right)\right)$

$\∫\frac{2}{3}\sqrt{-4{\sin \left(\mathrm{\θ}\right)}^2\+4}\cos \left(\mathrm{\θ}\right)\ⅆ\mathrm{\θ}$

$q_2≔\mathrm{simplify}\left(q_1\right) assuming \mathrm{\θ}\colon\colon \mathrm{RealRange}\left(-\frac{\mathrm{\π}}{2}\,\frac{\mathrm{\π}}{2}\right)$

$\frac{4}{3}\∫{\cos \left(\mathrm{\θ}\right)}^2\ⅆ\mathrm{\θ}$

$q_3≔\mathrm{simplify}\left(q_2\,\left\{{\cos }^2\left(\mathrm{\θ}\right)\=\left(1\+\cos \left(2 \mathrm{\θ}\right)\right)\/2\right\}\right)$

$\frac{4}{3}\∫\left(\frac{1}{2}\+\frac{1}{2}\cos \left(2\mathrm{\θ}\right)\right)\ⅆ\mathrm{\θ}$

$q_4≔\mathrm{expand}\left(q_3\,2 \mathrm{\θ}\right)$

$\frac{2}{3}\∫1\ⅆ\mathrm{\θ}\+\frac{2}{3}\∫\cos \left(2\mathrm{\θ}\right)\ⅆ\mathrm{\θ}$

$q_5≔\mathrm{value}\left(q_4\right)$

$\frac{2}{3}\mathrm{\θ}\+\frac{1}{3}\sin \left(2\mathrm{\θ}\right)$

$q_6≔\mathrm{expand}\left(q_5\right)$

$\frac{2}{3}\mathrm{\θ}\+\frac{2}{3}\sin \left(\mathrm{\θ}\right)\cos \left(\mathrm{\θ}\right)$

$\mathrm{eval}\left(q_6\,\mathrm{\θ}\=\arcsin \left(\frac{3}{2}x\right)\right)$

$\frac{1}{2}x\sqrt{-9x^2\+4}\+\frac{2}{3}\arcsin \left(\frac{3}{2}x\right)$

$\begin{array}{cccc}\multicolumn{4}{c}{\int \sqrt{-9x^2+4}\ⅆx}\\ & \text{=}& -\frac{4\int {\sin \left(u\right)}^2\ⅆu}{3}+\frac{4u}{3}\hfill & \left[\mathrm{change}\,x=\frac{2\sin \left(u\right)}{3}\,u\right]\\ & \text{=}& \frac{2u}{3}+\frac{2\int \cos \left(2u\right)\ⅆu}{3}\hfill & \left[\mathrm{rewrite}\,{\sin \left(u\right)}^2=\frac{1}{2}-\frac{\cos \left(2u\right)}{2}\right]\\ & \text{=}& \frac{2u}{3}+\frac{\int \cos \left(\mathrm{u1}\right)\ⅆ\mathrm{u1}}{3}\hfill & \left[\mathrm{change}\,\mathrm{u1}=2u\,\mathrm{u1}\right]\\ & \text{=}& \frac{2u}{3}+\frac{\sin \left(\mathrm{u1}\right)}{3}\hfill & \left[\cos \right]\\ & \text{=}& \frac{2u}{3}+\frac{\sin \left(2u\right)}{3}\hfill & \left[\mathrm{revert}\right]\\ & \text{=}& \frac{2\arcsin \left(\frac{3x}{2}\right)}{3}+\frac{x\sqrt{-9x^2+4}}{2}\hfill & \left[\mathrm{revert}\right]\end{array}$

Table 6.3.1(a)   Annotated stepwise solution via Integration Methods tutor