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

 

\begin{array}{l} \smallint \frac{{{{\rm{x}}^3}}}{{\sqrt {{{\left( {4{{\rm{x}}^2} + 9} \right)}^3}} }}\;dx\\\\ Let \Rightarrow 2 \cdot x = 3\tan \theta \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;\;\;2{\rm{d}}x = 3{\sec ^2}\theta {\rm{d}}\theta \\ Change{\rm{ }}the\;expression\\ 9 + 4{x^2} = 9{\tan ^2}\theta + 9\;\;\;\;\\ \;\;\;9 + 4{x^2} = 9\left( {{{\tan }^2}\theta + 1} \right)\;\;\;\;\\ \;\;\;9 + 4{x^2} = 9{\sec ^2}\theta \;\;\;\;\\ \;\;\;\sqrt {9 + 4{x^2}} = 3\sec \theta \;\;\;\;\\ \;\;\;{\left( {9 + 4{x^2}} \right)^{\left( {\frac{3}{2}} \right)}} = 27{\sec ^3}\theta \\ \end{array} \begin{array}{*{20}{c}} {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x = \frac{3}{2}\tan \theta }\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{x^3} = \frac{{27}}{8}{{\tan }^3}\theta } \end{array} \begin{array}{l} \smallint \frac{{\frac{{27}}{8}{{\tan }^3}\theta }}{{27{{\sec }^3}\theta }} \cdot \frac{3}{2} \cdot {\sec ^2}\theta {\rm{d}}\theta = \frac{3}{{16}}\smallint \frac{{ta{n^3}\theta }}{{sec\theta }}{\rm{d}}\theta \\ \frac{3}{{16}}\smallint \cos \theta \cdot \frac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}{\rm{d}}\theta = \frac{3}{{16}}\smallint \frac{{si{n^3}\theta }}{{co{s^2}\theta }}{\rm{d}}\theta \\ \frac{3}{{16}}\smallint \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \cdot \sin \theta {\rm{d}}\theta = \frac{3}{{16}}\smallint \frac{{1 - co{s^2}\theta }}{{co{s^2}\theta }} \cdot sin\theta {\rm{d}}\theta \\ - \frac{3}{{16}}\smallint \frac{{1 - {z^2}}}{{{z^2}}}{\rm{d}}z\\ - \frac{3}{{16}}\smallint {{\rm{z}}^{ - 2}} - 1{\rm{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 + 4{x^2}} }}{3} + \frac{3}{{\sqrt {9 + 4{x^2}} }}} \right) + c \end{array}