# Integrate x^3*√(16-x^2) by using Trigonometric Substitutions (only)???

May 1, 2018

$I = \frac{{\left(\sqrt{16 - {x}^{2}}\right)}^{5}}{5} - \frac{16 {\left(\sqrt{16 - {x}^{2}}\right)}^{3}}{3} + c$

#### Explanation:

Here,

$I = \int {x}^{3} \sqrt{16 - {x}^{2}} \mathrm{dx}$

Let,

$x = 4 \sin u \implies \mathrm{dx} = 4 \cos u \mathrm{du} , \mathmr{and}$

$\sin u = \frac{x}{4}$
$\implies \cos u = \sqrt{1 - {\sin}^{2} u} = \sqrt{1 - {x}^{2} / 16} = \frac{\sqrt{16 - {x}^{2}}}{4.} . . \to \left(1\right)$

So,

$I = \int 64 {\sin}^{3} u \sqrt{16 - 16 {\sin}^{2} u} \times 4 \cos u \mathrm{du}$

$= \int 64 {\sin}^{3} u \left(4 \cos u\right) 4 \cos u \mathrm{du}$

$= 1024 \int {\sin}^{3} u {\cos}^{2} u \mathrm{du}$

$= 1024 \int {\sin}^{2} u {\cos}^{2} u \sin u \mathrm{du}$

$= 1024 \int \left(1 - {\cos}^{2} u\right) {\cos}^{2} u \sin u \mathrm{du}$

$= 1024 \int \left[{\cos}^{2} u \sin u - {\cos}^{4} u \sin u\right] \mathrm{du}$

$= 1024 \left[- \int {\left(\cos u\right)}^{2} \left(- \sin u\right) \mathrm{du} + \int {\left(\cos u\right)}^{4} \left(- \sin u\right) \mathrm{du}\right]$

=1024[-int(cosu)^2(d/(dx)(cosu))du+int(cosu)^4(d/(dx)(cosu)du]

$= 1024 \left[- {\cos}^{3} \frac{u}{3} + {\cos}^{5} \frac{u}{5}\right] + c$

$= 1024 \left[{\left(\cos u\right)}^{5} / 5 - {\left(\cos u\right)}^{3} / 3\right] + c$

From $\left(1\right)$,we have $\cos u = \frac{\sqrt{16 - {x}^{2}}}{4}$

=4^5[(sqrt(16-x^2))^5/(4^5xx5)-(sqrt(16- x^2))^3/(4^3xx3)]+c

$I = \frac{{\left(\sqrt{16 - {x}^{2}}\right)}^{5}}{5} - \frac{16 {\left(\sqrt{16 - {x}^{2}}\right)}^{3}}{3} + c$

May 1, 2018

$I = \frac{1}{5} {\left(16 - {x}^{2}\right)}^{\frac{5}{2}} - \frac{16}{3} {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} + C$

#### Explanation:

Let $x = 4 \sin \theta$. Then $\mathrm{dx} = 4 \cos \theta d \theta$.

$I = \int {\left(4 \sin \theta\right)}^{3} \sqrt{16 - {\left(4 \sin \theta\right)}^{2}} d \theta \cdot 4 \cos \theta d \theta$

$I = \int 64 {\sin}^{3} \theta \left(4 \cos \theta\right) \left(4 \cos \theta\right) d \theta$

$I = 1024 \int {\sin}^{3} \theta {\cos}^{2} \theta d \theta$

$I = 1024 \int {\sin}^{2} \theta \sin \theta {\cos}^{2} \theta d \theta$

$I = 1024 \int \left(1 - {\cos}^{2} \theta\right) {\cos}^{2} \theta \sin \theta$

$I = 1024 \int {\cos}^{2} \theta \sin \theta - 1024 {\cos}^{4} \theta \sin \theta d \theta$

$I = 1024 \int {\cos}^{2} \theta \sin \theta d \theta - 1024 \int {\cos}^{4} \theta \sin \theta d \theta$

We now let $u = \cos \theta$. Then $\mathrm{du} = - \sin \theta d \theta$ and $d \theta = \frac{\mathrm{du}}{- \sin \theta}$.

$I = 1024 \int {u}^{2} \mathrm{du} - 1024 \int {u}^{4} \mathrm{du}$

$I = - 1024 \left(\frac{1}{3} {u}^{3}\right) + 1024 \left(\frac{1}{5} {u}^{5}\right) + C$

$I = - \frac{1024}{3} {u}^{3} + \frac{1024}{5} {u}^{5} + C$

$I = - \frac{1024}{3} {\cos}^{3} \theta + \frac{1024}{5} {\cos}^{5} \theta + C$

From our initial substitution we see that $\frac{x}{4} = \sin \theta$. Therefore, $\frac{\sqrt{16 - {x}^{2}}}{4} = \cos \theta$.

$I = - \frac{1024}{3} {\left(\frac{\sqrt{16 - {x}^{2}}}{4}\right)}^{3} + \frac{1024}{5} {\left(\frac{\sqrt{16 - {x}^{2}}}{4}\right)}^{5} + C$

$I = - \frac{16}{3} {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{1}{5} {\left(16 - {x}^{2}\right)}^{\frac{5}{2}} + C$

Hopefully this helps!