# How do you find the integral intx*sqrt(1-x^4)dx ?

Jul 27, 2014

Answer, $= \frac{1}{4} \left({\sin}^{- 1} \left({x}^{2}\right) + {x}^{2} \sqrt{1 - {x}^{4}}\right) + c$

Explanation :
$\int x \cdot \sqrt{1 - {x}^{4}} \mathrm{dx} = \int x \cdot \sqrt{1 - {\left({x}^{2}\right)}^{2}} \mathrm{dx}$

Using Trigonometric Substitution
let's ${x}^{2} = \sin t , \implies 2 x \mathrm{dx} = \cos t \mathrm{dt}$. Here I am using $\sin t$, it can also be done by considering ${x}^{2} = \cos t$.

Now plugging in integral,

$= \int \frac{1}{2} \sqrt{1 - {x}^{4}} \left(2 x \mathrm{dx}\right)$

$= \int \frac{1}{2} \cdot \cos t \cdot \sqrt{1 - {\sin}^{2} t} \mathrm{dt}$

Using the identity, ${\sin}^{2} t + {\cos}^{2} t = 1$
we get,

$= \int \frac{1}{2} \cdot {\cos}^{2} t \mathrm{dt}$,

Using another important identity, $\cos 2 t = {\cos}^{2} t - {\sin}^{2} t$
$\implies \cos 2 t = 2 {\cos}^{2} t - 1$

$\implies {\cos}^{2} t = \frac{1 + \cos 2 t}{2}$

$= \int \frac{1}{2} \cdot \frac{1}{2} \cdot \left(1 + \cos 2 t\right) \mathrm{dt} = \frac{1}{4} \int 1 \mathrm{dt} + \frac{1}{4} \int \cos 2 t \mathrm{dt}$

Using Trigonometric functions,

$= \frac{1}{4} \cdot t + \frac{1}{4} \frac{\sin 2 t}{2} + c$, where c is constant

$= \frac{1}{4} \cdot \left(t + \frac{\sin 2 t}{2}\right) + c$

Use the identity $\sin 2 t = 2 \sin t \cos t$:

$= \frac{1}{4} \left(t + \frac{2 \sin t \cos t}{2}\right) + c$
$= \frac{1}{4} \left(t + \sin t \cos t\right) + c$
$= \frac{1}{4} \left(t + \sin t \sqrt{1 - {\sin}^{2} t}\right) + c$

And substituting $\sin t = {x}^{2}$ back in:
$= \frac{1}{4} \left({\sin}^{- 1} \left({x}^{2}\right) + {x}^{2} \sqrt{1 - {x}^{4}}\right) + c$