# How do you integrate int x^3sqrt(4-9x^2) by trigonometric substitution?

May 12, 2018

I=1/1215[3sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c

#### Explanation:

Here,

$I = \int {x}^{3} \sqrt{4 - 9 {x}^{2}} \mathrm{dx}$

let,

$3 x = 2 \sin u \implies x = \frac{2}{3} \sin u \implies \mathrm{dx} = \frac{2}{3} \cos u \mathrm{du} \mathmr{and}$

$9 {x}^{2} = 4 {\sin}^{2} u \mathmr{and} \sin u = \frac{3 x}{2} \implies {\cos}^{2} u = 1 - \frac{9 {x}^{2}}{4}$

$\implies {\cos}^{2} u = \frac{4 - 9 {x}^{2}}{4} \implies \cos u = \frac{1}{2} \sqrt{4 - 9 {x}^{2}}$

So,

$I = \int {\left(\frac{2}{3} \sin u\right)}^{3} \sqrt{4 - 4 {\sin}^{2} u} \times \frac{2}{3} \cos u \mathrm{du}$

$= \int \frac{8}{27} {\sin}^{3} u \left(2 \cos u\right) \frac{2}{3} \cos u \mathrm{du}$

$= \frac{32}{81} \int {\sin}^{3} u {\cos}^{2} u \mathrm{du}$

$= \frac{32}{81} \int {\sin}^{2} u {\cos}^{2} u \sin u \mathrm{du}$

=-32/81int[(1-cos^2u)cos^2u(-sinu)du

$= - \frac{32}{81} \left[\int {\cos}^{2} u \left(- \sin u\right) \mathrm{du} - \int {\cos}^{4} u \left(- \sin u\right) \mathrm{du}\right]$

$= - \frac{32}{81} \left[\int {\left(\cos u\right)}^{2} \frac{d}{\mathrm{dx}} \left(\cos u\right) \mathrm{du} - \int {\left(\cos u\right)}^{4} \frac{d}{\mathrm{dx}} \left(\cos u\right) \mathrm{du}\right]$

$= - \frac{32}{81} \left[{\left(\cos u\right)}^{3} / 3 - {\left(\cos u\right)}^{5} / 5\right] + c$

$= - \frac{32}{81 \times 15} \left[5 {\left(\cos u\right)}^{3} - 3 {\left(\cos u\right)}^{5}\right] + c$

Subst. cosu=1/2sqrt(4-9x^2

I=-32/1215[5xx(sqrt(4-9x^2))^3/(8)-3xx(sqrt(4- 9x^2))^5/(32)]+c

$I = - \frac{1}{1215} \left[20 {\left(\sqrt{4 - 9 {x}^{2}}\right)}^{3} - 3 {\left(\sqrt{4 - 9 {x}^{2}}\right)}^{5}\right] + c$

I=1/1215[3sqrt(4-9x^2))^5-20(sqrt(4-9x^2))^3]+c

May 12, 2018

$I = \frac{1}{1215} \left[3 {\left(\sqrt{4 - 9 {x}^{2}}\right)}^{5} - 20 {\left(\sqrt{4 - 9 {x}^{2}}\right)}^{3}\right] + c$

#### Explanation:

Method Without trigonometric substitution.

$I = \int {x}^{3} \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = \int {x}^{2} \sqrt{4 - 9 {x}^{2}} \cdot x \mathrm{dx}$

Let,

$\sqrt{4 - 9 {x}^{2}} = u \implies 4 - 9 {x}^{2} = {u}^{2} \implies 9 {x}^{2} = 4 - {u}^{2}$

$\implies {x}^{2} = \frac{1}{9} \left(4 - {u}^{2}\right) \implies 2 x \mathrm{dx} = - \frac{2}{9} u \mathrm{du}$

$\implies x \mathrm{dx} = - \frac{1}{9} u \mathrm{du} \mathmr{and}$

So,

$I = \int \frac{1}{9} \left(4 - {u}^{2}\right) \left(u\right) \left(- \frac{1}{9} u\right) \mathrm{du}$

$= \frac{1}{81} \int \left({u}^{2} - 4\right) {u}^{2} \mathrm{du}$

$= \frac{1}{81} \int \left({u}^{4} - 4 {u}^{2}\right) \mathrm{du}$

$= \frac{1}{81} \left[{u}^{5} / 5 - \frac{4 {u}^{3}}{3}\right] + c$

$= \frac{1}{81 \times 15} \left[3 {u}^{5} - 20 {u}^{3}\right] + c$

Subst. $u = \sqrt{4 - 9 {x}^{2}}$

$I = \frac{1}{1215} \left[3 {\left(\sqrt{4 - 9 {x}^{2}}\right)}^{5} - 20 {\left(\sqrt{4 - 9 {x}^{2}}\right)}^{3}\right] + c$