$\int x \sqrt{4-9 x^2}\mathit{\ⅆ}x$

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

$\= \frac{8}{9}\int \sin \left(\mathrm{\θ}\right){\cos }^2\left(\mathrm{\θ}\right) \mathit{\ⅆ}\mathrm{\θ}$

$\= -\frac{8}{9}\frac{{\cos }^3\left(\mathrm{\θ}\right)}{3}$

$\= -\frac{8}{27}{\left(\frac{1}{2}\sqrt{4-9 x^2}\right)}^3$

$\= -\frac{1}{27}{\left(4-9 x^2\right)}^{3\/2}$

Evaluate the given integral

$$$\int x \sqrt{4-9 x^2}\mathit{\ⅆ}x$ = $\frac{1}{27}\left(3x-2\right)\left(3x\+2\right)\sqrt{-9x^2\+4}$$\stackrel{\text{simplify}}{\=}$$-\frac{1}{27}{\left(-9x^2\+4\right)}^{3\/2}$$$

Initialization

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

  $Q≔\int x \sqrt{4-9 x^2}\mathit{\ⅆ}x\:$

Change variables as per Table 6.3.1

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

$\∫\frac{4}{9}\sin \left(\mathrm{\θ}\right)\sqrt{-4{\sin \left(\mathrm{\θ}\right)}^2\+4}\cos \left(\mathrm{\θ}\right)\ⅆ\mathrm{\θ}$

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

$\frac{8}{9}\∫\sin \left(\mathrm{\θ}\right){\cos \left(\mathrm{\θ}\right)}^2\ⅆ\mathrm{\θ}$

$q_3≔\mathrm{value}\left(q_2\right)$

$-\frac{8}{27}{\cos \left(\mathrm{\θ}\right)}^3$

$\mathrm{eval}\left(q_3\,\mathrm{\θ}\=\arcsin \left(\frac{3}{2}x\right)\right)$

$-\frac{1}{27}{\left(-9x^2\+4\right)}^{3\/2}$

$\begin{array}{cccc}\multicolumn{4}{c}{\int x\sqrt{-9x^2+4}\ⅆx}\\ & \text{=}& -\frac{\int u^2\ⅆu}{9}\hfill & \left[\mathrm{change}\,-9x^2+4=u^2\,u\right]\\ & \text{=}& -\frac{u^3}{27}\hfill & \left[\mathrm{power}\right]\\ & \text{=}& -\frac{{\left(-9x^2+4\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{27}\hfill & \left[\mathrm{revert}\right]\end{array}$

Table 6.3.3(a)   Annotated stepwise solution via Integration Methods tutor