Two-step Radical

Since we know how to integrate $\sqrt{1-u^2}$ using a trigonometric substitution , we can use the substitution $u = x^2$ (and $du = 2x\,dx$ ) in order to simplify the integral:

$\int_0^1 x \sqrt{1 - x^4} \, dx = \int_0^1 \tfrac{1}{2} \sqrt{1-(x^2)^2} \cdot 2x \, dx = \int_0^1 \tfrac{1}{2} \sqrt{1 - u^2}\, du.$

Now, we can use the trigonometric substitution $u = \sin\theta$ (and $du = \cos\theta \, d\theta$ ) to obtain:

$\int_0^1 \tfrac{1}{2} \sqrt{1 - u^2}\, du = \int_0^{\pi/2} \tfrac{1}{2} \sqrt{1 - \sin^2\theta} \cos\theta \, d\theta = \int_0^{\pi/2} \tfrac{1}{2} \cos^2\theta \, d\theta.$

Finally, using the half-angle identity $\cos^2\theta = \tfrac{1 + \cos(2\theta)}{2}$ , the integral evaluates to:

$\int_0^{\pi/2} \tfrac{1}{2} \cos^2\theta \, d\theta = \int_0^{\pi/2} \frac{1}{4} + \frac{\cos(2\theta)}{4} \, d\theta = \left[ \frac{x}{4} + \frac{\sin(2\theta)}{8} \right]_0^{\pi/2} = \boxed{\frac{\pi}{8}.}$

$\begin{aligned} I & = \int_0^1 x\sqrt{1-\color{#3D99F6}{x^4}} dx \quad \quad \small \color{#3D99F6}{\text{Let }u = x^2 \implies du = 2x \ dx} \\ & = \frac12 \int_0^1 \sqrt{1-u^2} du \\ & = \frac12 \int_0^1 u^0 (1-u^2)^\frac12 du \\ & = \frac14\color{#3D99F6}{B \left(\frac12, \frac32 \right)} \quad \quad \small \color{#3D99F6}{B(m,n) \text{ is beta function}} \\ & = \frac{\color{#3D99F6}{\Gamma \left(\frac12 \right) \Gamma \left(\frac32 \right)}}{4\color{#3D99F6}{\Gamma \left( 2 \right)}} \quad \quad \small \color{#3D99F6}{\Gamma (n) \text{ is gamma function}} \\ & = \frac{\frac12 \left(\Gamma \left(\frac12 \right) \right)^2}{4\cdot 1!} \\ & = \frac{(\sqrt{\pi})^2}{8} = \boxed{\dfrac\pi 8} \end{aligned}$