# Question #e678e

Dec 5, 2017

${\int}_{0}^{1} x \cdot \sqrt{1 - {x}^{4}} \cdot \mathrm{dx} = \frac{\pi}{8}$

#### Explanation:

${\int}_{0}^{1} x \cdot \sqrt{1 - {x}^{4}} \cdot \mathrm{dx}$

=$\frac{1}{2} \cdot {\int}_{0}^{1} \sqrt{1 - {\left({x}^{2}\right)}^{2}} \cdot 2 x \cdot \mathrm{dx}$

After using $\sin u = {x}^{2}$ and $\cos u \cdot \mathrm{du} = 2 x \cdot \mathrm{dx}$ substitution, this integral became

$\frac{1}{2} \cdot {\int}_{0}^{\frac{\pi}{2}} \sqrt{1 - {\left(\sin u\right)}^{2}} \cdot \cos u \cdot \mathrm{du}$

=$\frac{1}{2} \cdot {\int}_{0}^{\frac{\pi}{2}} \sqrt{{\left(\cos u\right)}^{2}} \cdot \cos u \cdot \mathrm{du}$

=$\frac{1}{2} \cdot {\int}_{0}^{\frac{\pi}{2}} {\left(\cos u\right)}^{2} \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\int}_{0}^{\frac{\pi}{2}} \left(1 + \cos 2 u\right) \cdot \mathrm{du}$

=${\left[\frac{u}{4} + \frac{1}{8} \cdot \sin 2 u\right]}_{0}^{\frac{\pi}{2}}$

=$\frac{\pi}{8}$

Dec 5, 2017

$\frac{\pi}{8}$

#### Explanation:

.

${\int}_{0}^{1} x \sqrt{1 - {x}^{4}} \mathrm{dx}$

$u = {x}^{2}$
$x = \sqrt{u}$
${u}^{2} = {x}^{4}$
$\mathrm{du} = 2 x \mathrm{dx}$
$\mathrm{dx} = \frac{\mathrm{du}}{2 x} = \frac{\mathrm{du}}{2 \sqrt{u}}$
$\sqrt{1 - {x}^{4}} = \sqrt{1 - {u}^{2}}$

Now, we substitute the above into the problem integral to turn it into an integral in terms of $u$:

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

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

Now we can use trigonometric substitution to solve this:

$u = \sin \theta$
$\mathrm{du} = \cos \theta d \left(\theta\right)$

$\frac{1}{2} \int \sqrt{1 - {\sin}^{2} \theta} \cos \theta d \left(\theta\right)$

$\frac{1}{2} \int \sqrt{{\cos}^{2} \theta} \cos \theta d \left(\theta\right)$

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

${\cos}^{2} \theta = \frac{1 + \cos \left(2 \theta\right)}{2}$

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

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

Now, we can substitute back for $u$:

$u = \sin \theta$, then $\theta = \arcsin u$
$\cos \theta = \sqrt{1 - s i {x}^{2} \theta} = \sqrt{1 - {u}^{2}}$
$\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta = 2 u \sqrt{1 - {u}^{2}}$

$\frac{1}{4} \theta + \frac{1}{8} \sin \left(2 \theta\right) = \frac{\arcsin u + u \sqrt{1 - {u}^{2}}}{4}$

Now, we can substitute back for $x$:

${\int}_{0}^{1} x \sqrt{1 - {x}^{4}} \mathrm{dx} = \frac{\arcsin {x}^{2} + {x}^{2} \sqrt{1 - {x}^{4}}}{4}$

We will evaluate this from $0$ to $1$:

$\frac{\arcsin {1}^{2} + {1}^{2} \sqrt{1 - {1}^{4}}}{4} - \frac{\arcsin {0}^{2} + {0}^{2} \sqrt{1 - {0}^{4}}}{4} =$

$\frac{\frac{\pi}{2} + 0}{4} - \frac{0 + 0}{4} = \frac{\pi}{8}$