# How do you integrate int x sqrt( 1 - x^4 )dx using trigonometric substitution?

Mar 12, 2018

If we let $u = {x}^{2}$, then $\mathrm{du} = 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{2 x}$.

$I = \int x \sqrt{1 - {u}^{2}} \frac{\mathrm{du}}{2 x}$

$I = \frac{1}{2} \int \sqrt{1 - {u}^{2}} \mathrm{du}$

This is when the trig substitution comes in. Let $u = \sin \theta$. Then $\mathrm{du} = \cos \theta d \theta$.

$I = \frac{1}{2} \int \sqrt{1 - {\sin}^{2} \theta} \cos \theta d \theta$

$I = \frac{1}{2} \int \sqrt{{\cos}^{2} \theta} \cos \theta d \theta$

$I = \frac{1}{2} \int \cos \theta \cos \theta d \theta$

$I = \frac{1}{2} \int {\cos}^{2} \theta d \theta$

We know that $\cos \left(2 \theta\right) = 2 {\cos}^{2} \theta - 1$, so $\frac{1}{2} \cos \left(2 \theta\right) = {\cos}^{2} \theta - \frac{1}{2}$ and $\frac{1}{2} \cos \left(2 \theta\right) + \frac{1}{2} = {\cos}^{2} \theta$. This is the famous power reduction identity, frequently rewritten as $\frac{1}{2} \left(\cos 2 \theta + 1\right)$

$I = \frac{1}{2} \int \frac{1}{2} \left(\cos 2 \theta + 1\right) d \theta$

$I = \frac{1}{4} \int \cos \left(2 \theta\right) + 1 d \theta$

$I = \frac{1}{4} \left(\frac{1}{2} \sin \left(2 \theta\right)\right) + \frac{1}{4} \theta + C$

$I = \frac{1}{8} \sin \left(2 \theta\right) + \frac{1}{4} \theta + C$

Recall that $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$.

$I = \frac{1}{8} \left(2\right) \sin \theta \cos \theta + \frac{1}{4} \theta + C$

$I = \frac{1}{4} \sin \theta \cos \theta + \frac{1}{4} \theta + C$

From our initial $\theta$-substitution, we know that $\sin \theta = \frac{u}{1}$. Thus $\theta = \arcsin \left(u\right)$ and $\cos \theta = \sqrt{1 - {u}^{2}}$

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

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

Hopefully this helps!