# How do you integrate int xsqrt(1-x^4) by trigonometric substitution?

Sep 6, 2016

$\frac{{x}^{2} \sqrt{1 - {x}^{4}} - \arcsin \left({x}^{2}\right)}{4} + C$

#### Explanation:

We have:

$I = \int x \sqrt{1 - {x}^{4}} \mathrm{dx}$

Let ${x}^{2} = \sin \left(u\right)$. This implies that $2 x \mathrm{dx} = \cos \left(u\right) \mathrm{du}$. So, we have:

$I = \frac{1}{2} \int 2 x \sqrt{1 - {\left({x}^{2}\right)}^{2}} \mathrm{dx}$

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

Since ${\sin}^{2} \left(u\right) + {\cos}^{2} \left(u\right) = 1$, we see that cos(u)=sqrt(1-sin^2(u):

$I = \frac{1}{2} \int {\cos}^{2} \left(u\right) \mathrm{du}$

To integrate this, recall that $\cos \left(2 u\right) = 2 {\cos}^{2} \left(u\right) - 1$, so ${\cos}^{2} \left(u\right) = \frac{1}{2} \left(\cos \left(2 u\right) - 1\right)$. Plugging this into the integral:

$I = \frac{1}{4} \int \left(\cos \left(2 u\right) - 1\right) \mathrm{du}$

$I = \frac{1}{4} \int \cos \left(2 u\right) \mathrm{du} - \frac{1}{4} \int \mathrm{du}$

Solving both of these integrals:

$I = \frac{1}{8} \int 2 \cos \left(2 u\right) \mathrm{du} - \frac{1}{4} u$

$I = \frac{1}{8} \sin \left(2 u\right) - \frac{1}{4} u$

$I = \frac{1}{8} \left(2 \sin \left(u\right) \cos \left(u\right)\right) - \frac{1}{4} u$

$I = \frac{1}{4} \sin \left(u\right) \cos \left(u\right) - \frac{1}{4} u$

Write the cosine function as sine, since we are working with a sine-based substitution:

$I = \frac{1}{4} \sin \left(u\right) \sqrt{1 - {\sin}^{2} \left(u\right)} - \frac{1}{4} u$

Plugging in $\sin \left(u\right) = {x}^{2}$ and $u = \arcsin \left({x}^{2}\right)$:

$I = \frac{1}{4} \left({x}^{2}\right) \sqrt{1 - {\left({x}^{2}\right)}^{2}} - \frac{1}{4} \arcsin \left({x}^{2}\right)$

$I = \frac{{x}^{2} \sqrt{1 - {x}^{4}} - \arcsin \left({x}^{2}\right)}{4} + C$