How to Integrate x^3/sqrt[(4x^2+9)^3] by Trigonometric Substitution
Published On :2021-01-28 20:31:00
begin{array}{l} smallint frac{{{{rm{x}}^3}}}{{sqrt {{{left( {4{{rm{x}}^2} + 9} right)}^3}} }};dx\\ Let Rightarrow 2 cdot x = 3tan 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} = 9left( {{{tan }^2}theta + 1} right);;;;\ ;;;9 + 4{x^2} = 9{sec ^2}theta ;;;;\ ;;;sqrt {9 + 4{x^2}} = 3sec 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 }}{{sectheta }}{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 sintheta {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}
