SOLUTION 5: $\ \$ To integrate $\displaystyle{ \int{x^3 \sqrt{4-9x^2} } \ dx } = \displaystyle{ \int {x^3 \sqrt{4(1-(9/4)x^2)} } \ dx } = \displaystyle{ 2 \int {x^3 \sqrt{1-((3/2)x)^2} } \ dx }$ use the trig substitution $$x = (2/3)\sin \theta$$ so that $$dx = (2/3) \cos \theta \ d \theta$$ Substitute into the original problem, replacing all forms of x, getting $$\displaystyle {\displaystyle{ 2 \int {x^3 \sqrt{1-((3/2)x)^2} } \ dx } = 2 \int {( (2/3)\sin \theta )^3 \sqrt{1-((3/2)( (2/3)\sin \theta ) )^2} } (2/3) \cos \theta \ d \theta }$$ $$= \displaystyle { 4/3 \int { (2/3)^3 \sin^3 \theta \sqrt{1-\sin^2 \theta } } \cos \theta \ d \theta }$$ $$= \displaystyle { 32/81 \int { \sin^3 \theta \cos \theta \cos \theta } \ d \theta }$$ $$= \displaystyle { 32/81 \int { \sin^3 \theta \cos^2 \theta } \ d \theta }$$ (Recall that $\sin^2 \theta = 1 - \cos^2 \theta$.) $$= \displaystyle { 32/81 \int { \sin \theta \sin^2 \theta \cos^2 \theta } \ d \theta }$$ $$= \displaystyle { 32/81 \int { \sin \theta ( 1 - \cos^2 \theta ) \cos^2 \theta } \ d \theta }$$ $$= \displaystyle { 32/81 \int { \sin \theta ( \cos^2 \theta - \cos^4 \theta) } \ d \theta }$$ (Now let $u= \cos \theta \ \longrightarrow \ du= - \sin \theta \ \longrightarrow \ -du = \sin \theta$.) $$= \displaystyle { - 32/81 \int { (u^2 - u^4) } \ d \theta }$$ $$= \displaystyle { - \frac{32}{81} { \Big( \frac{ u^3}{ 3 } - \frac{ u^5 }{5} \Big) } + C }$$ $$= \displaystyle { - \frac{32}{81} { \Big( \frac{ (\cos \theta )^3}{ 3 } - \frac{ ( \cos \theta )^5 }{5} \Big) } + C }$$ $\Big($We need to write our final answer in terms of $x$.

Since $x = (2/3) \sin \theta$ it follows that $$\sin \theta = \displaystyle{ 3x \over 2 } = \displaystyle{ opposite \over hypotenuse }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (3x)^2 = (2)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{4-9x^2} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{4-9x^2} \over 2 } . \Big)$$ $$= \displaystyle { - \frac{32}{81} \Big( \Big( \frac{1}{3} \Big( { \sqrt{4-9x^2} \over 2 } \Big)^3 - \frac{1}{5} \Big( { {\sqrt{4-9x^2} \over 2 } \Big)^5 } \Big) + C }$$ $$= \displaystyle { - \frac{32}{81} \Big( \frac{1}{3} \cdot \frac{1}{8} (4-9x^2)^{3/2} - \frac{1}{5} \cdot \frac{1}{32} (4-9x^2)^{5/2} \Big) + C }$$ $$= \displaystyle { - \frac{4}{243} (4-9x^2)^{3/2} + \frac{1}{405} (4-9x^2)^{5/2} + C }$$