# \int_(0)^(15)x^2\sqrt(a^2-x^2)dx?

## I got as far as $\setminus \int {a}^{4} \setminus {\sin}^{4} \left(\setminus \theta\right) \setminus {\cos}^{2} \left(\theta\right) d \setminus \theta$ from trigonometric substitution with $x = a \setminus \sin \left(\setminus \theta\right)$ $\mathrm{dx} = a \setminus \cos \left(\setminus \theta\right) d \setminus \theta$ NOTE The original integral looked like this $\setminus {\int}_{0}^{a} {x}^{2} \setminus \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx}$ In this version, the upper integration limit is $a$, not $15$.

Apr 15, 2018

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = \frac{{a}^{4} \pi}{16}$

#### Explanation:

Evaluate:

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx}$

Substitute:

$x = a \sin t$

$\mathrm{dx} = a \cos t \mathrm{dt}$

with $t \in \left[0 , \frac{\pi}{2}\right]$

so that:

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = {\int}_{0}^{\frac{\pi}{2}} {a}^{2} {\sin}^{2} t \sqrt{{a}^{2} - {a}^{2} {\sin}^{2} t} a \cos t \mathrm{dt}$

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = {a}^{4} {\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} t \sqrt{1 - {\sin}^{2} t} \cos t \mathrm{dt}$

For $t \in \left[0 , \frac{\pi}{2}\right]$ the cosine os positive, so:

$\sqrt{1 - {\sin}^{2} t} = \cos t$

and then:

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = {a}^{4} {\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} t {\cos}^{2} t \mathrm{dt}$

Use now the trigonometric identities:

$\sin 2 \theta = 2 \sin \theta \cos \theta$

$2 {\sin}^{2} \theta = 1 - \cos \theta$

so:

${\sin}^{2} t {\cos}^{2} t \mathrm{dt} = \frac{1}{4} {\sin}^{2} 2 t = \frac{1}{8} \left(1 - \cos 4 t\right)$

and:

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = {a}^{4} / 8 {\int}_{0}^{\frac{\pi}{2}} \left(1 - \cos 4 t\right) \mathrm{dt}$

Using now the linearity of the integral:

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = {a}^{4} / 8 \left({\int}_{0}^{\frac{\pi}{2}} \mathrm{dt} - {\int}_{0}^{\frac{\pi}{2}} \cos 4 t \mathrm{dt}\right)$

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = {a}^{4} / 8 {\left[t - \frac{\sin 4 t}{4}\right]}_{0}^{\frac{\pi}{2}}$

${\int}_{0}^{a} {x}^{2} \sqrt{{a}^{2} - {x}^{2}} \mathrm{dx} = \frac{{a}^{4} \pi}{16}$