How to Integrate x^3/sqrt[(4x^2+9)^3] by Trigonometric Substitution

 

$$\begin{array}{l}\int \frac{{\mathrm{x}}^3}{\sqrt{{\left(4{\mathrm{x}}^2+9\right)}^3}}dx\\ \\ Let\Rightarrow 2\cdot x=3\tan \theta \\ 2\mathrm{d}x=3{\sec }^2\theta \mathrm{d}\theta \\ Changetheexpression\\ 9+4x^2=9{\tan }^2\theta +9\\ 9+4x^2=9\left({\tan }^2\theta +1\right)\\ 9+4x^2=9{\sec }^2\theta \\ \sqrt{9+4x^2}=3\sec \theta \\ {\left(9+4x^2\right)}^{\left(\frac{3}{2}\right)}=27{\sec }^3\theta \end{array}$$ $$\begin{array}{c}x=\frac{3}{2}\tan \theta \\ x^3=\frac{27}{8}{\tan }^3\theta \end{array}$$ $$\begin{array}{l}\int \frac{\frac{27}{8}{\tan }^3\theta }{27{\sec }^3\theta }\cdot \frac{3}{2}\cdot {\sec }^2\theta \mathrm{d}\theta =\frac{3}{16}\int \frac{ta{n}^3\theta }{sec\theta }\mathrm{d}\theta \\ \frac{3}{16}\int \cos \theta \cdot \frac{{\sin }^3\theta }{{\cos }^3\theta }\mathrm{d}\theta =\frac{3}{16}\int \frac{si{n}^3\theta }{co{s}^2\theta }\mathrm{d}\theta \\ \frac{3}{16}\int \frac{{\sin }^2\theta }{{\cos }^2\theta }\cdot \sin \theta \mathrm{d}\theta =\frac{3}{16}\int \frac{1-co{s}^2\theta }{co{s}^2\theta }\cdot sin\theta \mathrm{d}\theta \\ -\frac{3}{16}\int \frac{1-z^2}{z^2}\mathrm{d}z\\ -\frac{3}{16}\int {\mathrm{z}}^{-2}-1\mathrm{d}z\\ -\frac{3}{16}\left(-\frac{1}{z}-z\right)\\ \Rightarrow \frac{3}{16}\left(\frac{1}{\cos \theta }+\cos \theta \right)+c\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{3}{16}\left(\frac{\sqrt{9+4x^2}}{3}+\frac{3}{\sqrt{9+4x^2}}\right)+c\end{array}$$