Question

Evaluate $\int \frac{(\sqrt[3]{x + \sqrt{2 - x^2}})(\sqrt[6]{1 - x\sqrt{2 - x^2}})dx}{\sqrt[3]{1 - x^2}}, x \in (0, 1):$

Answer 1

$2^{1/6} x + C$

Answer 2

To evaluate the integral $\int \frac{(\sqrt[3]{x + \sqrt{2 - x^2}})(\sqrt[6]{1 - x\sqrt{2 - x^2}})dx}{\sqrt[3]{1 - x^2}}$, for $x \in (0, 1)$, we use a trigonometric substitution.

1. Substitution: Given the term $\sqrt{2 - x^2}$, a suitable substitution is $x = \sqrt{2} \sin \theta$. Then $dx = \sqrt{2} \cos \theta d\theta$.

For $x \in (0, 1)$: $0 < \sqrt{2} \sin \theta < 1 \implies 0 < \sin \theta < \frac{1}{\sqrt{2}}$. This implies $\theta \in (0, \frac{\pi}{4})$.

2. Transform the terms in the integrand:

• $\sqrt{2 - x^2} = \sqrt{2 - (\sqrt{2} \sin \theta)^2} = \sqrt{2 - 2 \sin^2 \theta} = \sqrt{2(1 - \sin^2 \theta)} = \sqrt{2 \cos^2 \theta}$. Since $\theta \in (0, \frac{\pi}{4})$, $\cos \theta > 0$, so $\sqrt{2 \cos^2 \theta} = \sqrt{2} \cos \theta$.

• Numerator term 1: $\sqrt[3]{x + \sqrt{2 - x^2}} = \sqrt[3]{\sqrt{2} \sin \theta + \sqrt{2} \cos \theta} = \sqrt[3]{\sqrt{2}(\sin \theta + \cos \theta)} = (\sqrt{2})^{1/3} (\sin \theta + \cos \theta)^{1/3} = 2^{1/6} (\sin \theta + \cos \theta)^{1/3}$.

• Numerator term 2: $\sqrt[6]{1 - x\sqrt{2 - x^2}} = \sqrt[6]{1 - (\sqrt{2} \sin \theta)(\sqrt{2} \cos \theta)} = \sqrt[6]{1 - 2 \sin \theta \cos \theta} = \sqrt[6]{1 - \sin(2\theta)}$. We know that $1 - \sin(2\theta) = \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = (\cos \theta - \sin \theta)^2$. So, $\sqrt[6]{1 - \sin(2\theta)} = \sqrt[6]{(\cos \theta - \sin \theta)^2} = (\cos \theta - \sin \theta)^{2/6} = (\cos \theta - \sin \theta)^{1/3}$. Since $\theta \in (0, \frac{\pi}{4})$, $\cos \theta > \sin \theta$, so $\cos \theta - \sin \theta > 0$.

• Denominator term: $\sqrt[3]{1 - x^2} = \sqrt[3]{1 - (\sqrt{2} \sin \theta)^2} = \sqrt[3]{1 - 2 \sin^2 \theta} = \sqrt[3]{\cos(2\theta)}$. Since $\theta \in (0, \frac{\pi}{4})$, $2\theta \in (0, \frac{\pi}{2})$, so $\cos(2\theta) > 0$.

3. Substitute into the integral: The product of the numerator terms is: $2^{1/6} (\sin \theta + \cos \theta)^{1/3} \cdot (\cos \theta - \sin \theta)^{1/3}$ $= 2^{1/6} [(\sin \theta + \cos \theta)(\cos \theta - \sin \theta)]^{1/3}$ $= 2^{1/6} [\cos^2 \theta - \sin^2 \theta]^{1/3}$ $= 2^{1/6} [\cos(2\theta)]^{1/3}$.

Now, substitute all terms back into the integral: $I = \int \frac{2^{1/6} [\cos(2\theta)]^{1/3}}{[\cos(2\theta)]^{1/3}} (\sqrt{2} \cos \theta) d\theta$ $I = \int 2^{1/6} \cdot \sqrt{2} \cos \theta d\theta$ $I = \int 2^{1/6} \cdot 2^{1/2} \cos \theta d\theta$ $I = \int 2^{1/6 + 3/6} \cos \theta d\theta$ $I = \int 2^{4/6} \cos \theta d\theta$ $I = \int 2^{2/3} \cos \theta d\theta$ $I = 2^{2/3} \sin \theta + C$

4. Substitute back to $x$: From $x = \sqrt{2} \sin \theta$, we have $\sin \theta = \frac{x}{\sqrt{2}}$. $I = 2^{2/3} \left(\frac{x}{\sqrt{2}}\right) + C$ $I = 2^{2/3} \frac{x}{2^{1/2}} + C$ $I = 2^{2/3 - 1/2} x + C$ $I = 2^{(4-3)/6} x + C$ $I = 2^{1/6} x + C$

The final answer is $\boxed{2^{1/6} x + C}$.

Explanation of the solution:

The integral is simplified by the substitution $x = \sqrt{2} \sin \theta$. This transforms terms involving $\sqrt{2-x^2}$ into trigonometric functions. Key identities used are $1 - 2\sin^2\theta = \cos(2\theta)$ and $1 - \sin(2\theta) = (\cos\theta - \sin\theta)^2$. The terms in the numerator combine to $2^{1/6}(\cos(2\theta))^{1/3}$, which cancels with the denominator. The integral then simplifies to $2^{2/3} \int \cos\theta d\theta = 2^{2/3} \sin\theta + C$. Substituting back $\sin\theta = x/\sqrt{2}$ yields the final result $2^{1/6} x + C$.