# How do you integrate int x^3sqrt(16-x^2) by trigonometric substitution?

May 5, 2018

$\int {x}^{3} \sqrt{16 - {x}^{2}} \mathrm{dx} = - \frac{16}{3} {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{1}{5} {\left(16 - {x}^{2}\right)}^{\frac{5}{2}} + c$

#### Explanation:

Let $x = 4 \sin t$, then $\sqrt{16 - {x}^{2}} = 4 \cos t$ and $\mathrm{dx} = 4 \cos t \mathrm{dt}$

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

= $\int 64 {\sin}^{3} t \cdot 4 \cos t \cdot 4 \cos t \mathrm{dt}$

= $1024 \int {\sin}^{3} t {\cos}^{2} t \mathrm{dt}$

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

= $1024 \int \left({\cos}^{2} t - {\cos}^{4} t\right) \sin t \mathrm{dt}$

Let $\cos t = u$ then $\mathrm{du} = - \sin t \mathrm{dt}$ and our integral becomes

$- 1024 \int \left({u}^{2} - {u}^{4}\right) \mathrm{du}$

= $- 1024 \left({u}^{3} / 3 - {u}^{5} / 5\right)$

i.e. $1024 \left(- \frac{1}{3} {\cos}^{3} t + \frac{1}{5} {\cos}^{5} t\right)$ (substituting $u = \cos t$)

and as $\cos t = \frac{1}{4} \sqrt{16 - {x}^{2}}$, our integral becomes

$1024 \left(- \frac{1}{3} \cdot \frac{1}{64} {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} + \frac{1}{5} \cdot \frac{1}{1024} {\left(16 - {x}^{2}\right)}^{\frac{5}{2}}\right) + c$

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

May 5, 2018

Let,$\sqrt{16 - {x}^{2}} = t \implies 16 - {x}^{2} = {t}^{2} \implies x \mathrm{dx} = - t \mathrm{dt}$
$I = \int \left(16 - {t}^{2}\right) t \left(- t\right) \mathrm{dt} = \int \left({t}^{4} - 16 {t}^{2}\right) \mathrm{dt} = {t}^{5} / 5 - \frac{16 {t}^{3}}{3} + c$
Substituting, $t = \sqrt{16 - {x}^{2}}$
$I = {\left(\sqrt{16 - {x}^{2}}\right)}^{5} / 5 - \frac{16 {\left(\sqrt{16 - {x}^{2}}\right)}^{3}}{3} + c$

#### Explanation:

Here,

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

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

and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16

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

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

$= 256 \int {\sin}^{3} u \times 4 \cos u \times \cos u \mathrm{du}$

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

$I = {4}^{5} \int \left(1 - {\cos}^{2} u\right) {\cos}^{2} u \times \sin u \mathrm{du}$

$= {4}^{5} \left[\int {\cos}^{2} u \sin u \mathrm{du} - \int {\cos}^{4} u \sin u \mathrm{du}\right]$

$= {4}^{5} \left[- \int {\left(\cos u\right)}^{2} \left(- \sin u\right) \mathrm{du} + \int {\left(\cos u\right)}^{4} \left(- \sin u\right) \mathrm{du}\right]$

$= {4}^{5} \left[- {\left(\cos u\right)}^{3} / 3 + {\left(\cos u\right)}^{5} / 5\right] + c$

$= {4}^{5} \left[{\left(\cos u\right)}^{5} / 5 - {\left(\cos u\right)}^{3} / 3\right] + c \ldots \to$from $\left(A\right)$

$= {4}^{5} \left[{\left(\sqrt{16 - {x}^{2}}\right)}^{5} / \left(5 \times {4}^{5}\right) - {\left(\sqrt{16 - {x}^{2}}\right)}^{3} / \left(3 \times {4}^{3}\right)\right] + c$

$= {\left(\sqrt{16 - {x}^{2}}\right)}^{5} / 5 - {4}^{2} \times {\left(\sqrt{16 - {x}^{2}}\right)}^{3} / 3 + c$

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