# How do you integrate int x^3 /sqrt(4 - 2x^2) dx using trigonometric substitution?

Feb 7, 2016

$- \frac{3}{2} \sqrt{1 - {x}^{2} / 2} + \frac{1}{6} \sqrt{1 - {\left(\frac{3}{\sqrt{2}} x - \sqrt{2} {x}^{2}\right)}^{2}} + C$

#### Explanation:

Try the substitution: $x = \sqrt{2} \sin \left(u\right)$

This would mean $\mathrm{dx} = \sqrt{2} \cos \left(u\right) \mathrm{dx}$

Now put this into the integral and we get:

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

Tidying this up a little and using the trig -identity:
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, we get:

$\int \frac{4 {\sin}^{3} \left(u\right)}{2 \sqrt{1 - {\sin}^{2} \left(x\right)}} \cos \left(u\right) \mathrm{du}$

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

At this point we need another trig identity . The trig identity that we will use is: ${\sin}^{3} x = \frac{3}{4} \sin x - \frac{1}{4} \sin \left(3 x\right)$.

Putting that into the integral we get:

$2 \int \frac{3}{4} \sin u - \frac{1}{4} \sin \left(3 u\right) \mathrm{dx}$

Which can now be integrated to obtain:

$- \frac{3}{2} \cos \left(u\right) + \frac{1}{6} \cos \left(3 u\right) + C$

We obviously have the challenge now of reversing the substituting. Again using the trig identity near the start, this formula can be re - arranged to give:

$- \frac{3}{2} \sqrt{1 - {\sin}^{2} \left(u\right)} + \frac{1}{6} \sqrt{1 - {\sin}^{2} \left(3 u\right)} + C$

At this point we can now rearrange the above trig identity to get:

$\sin \left(3 u\right) = 3 \sin \left(u\right) - 4 {\sin}^{3} \left(u\right)$

We can now rewrite the above expression as:

$- \frac{3}{2} \sqrt{1 - {\sin}^{2} \left(u\right)} + \frac{1}{6} \sqrt{1 - {\left(3 \sin \left(u\right) - 4 {\sin}^{3} \left(u\right)\right)}^{2}} + C$

And finally we can reverse the substitution to obtain:

$- \frac{3}{2} \sqrt{1 - {x}^{2} / 2} + \frac{1}{6} \sqrt{1 - {\left(\frac{3}{\sqrt{2}} x - \sqrt{2} {x}^{2}\right)}^{2}} + C$