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

Apr 7, 2018

Sometimes trig.substitution is not easy. See and compare.
Let,$16 - {x}^{2} = {t}^{2} \implies x \mathrm{dx} = - t \mathrm{dt}$
$I = - \int \left(16 - {t}^{2}\right) {t}^{2} \mathrm{dt} = \int \left({t}^{4} - 16 {t}^{2}\right) \mathrm{dt} = {t}^{5} / 5 - \frac{16 {t}^{3}}{3} + c$
$I = {\left(\sqrt{16 - {x}^{2}}\right)}^{5} / 5 - \frac{16 {\left(\sqrt{16 - {x}^{2}}\right)}^{3}}{3} + C$

#### Explanation:

Here,

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

Let, $x = 4 \sin u \implies {x}^{2} = 16 {\sin}^{2} u \implies 2 x \mathrm{dx} = 32 \sin u \cos u \mathrm{du}$

i.e. $x \mathrm{dx} = 16 \sin u \cos u \mathrm{du}$

$\implies I = \int \left(16 {\sin}^{2} u\right) \sqrt{16 - 16 {\sin}^{2} u} \left(16 \sin u \cos u\right) \mathrm{du}$

$I = 16 \times 4 \times 16 \int {\sin}^{2} u \left(\cos u\right) \sin u \cos u \mathrm{du}$

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

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

=-1024[int(cosu)^2(-sinu)du-int(cosu)^4(-sinu)du

$= - 1024 \left[{\left(\cos u\right)}^{3} / 3 - {\left(\cos u\right)}^{5} / 5\right] + C \to w h e r e , {\sin}^{2} u = {x}^{2} / 16$

$= - 1024 \left[{\left(\sqrt{1 - {\sin}^{2} u}\right)}^{3} / 3 - {\left(\sqrt{1 - {\sin}^{2} u}\right)}^{5} / 5\right] + C$

$= - 1024 \left[{\left(\sqrt{1 - {x}^{2} / 16}\right)}^{3} / 3 - {\left(\sqrt{1 - {x}^{2} / 16}\right)}^{5} / 5\right] + C$

=-1024[(sqrt(16-x^2))^3/(3xx64)-(sqrt(16- x^2))^5/(5xx1024)]+C

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

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