Question

The value of the integral \[ \int\frac{\left(\sqrt[3]{x+\sqrt{2-x^{2}}}\right)\left(\sqrt[6]{1-x\sqrt{2-x^{2}}}\right)}{\sqrt[3]{1-x^{2}}}\,dx \] for \(x\in(0,1)\) is:

• A. $2^{\frac{1}{12}}x+c$
• B. $2^{\frac{3}{4}}x+c$
• C. $2^{\frac{1}{3}}x+c$
• D. $2^{\frac{1}{6}}x+c$

Hint:

Whenever you see nested radical functions with different roots (like a 3rd root and a 6th root) containing similar expressions, try to convert them to share a common root exponent first. This often reveals a hidden difference-of-squares cancellation that collapses the entire problem!

Answer 1

The Correct Option is A

Concept: Complex algebraic integrals containing radical combinations of the form $\sqrt{a^2-x^2}$ can be simplified by using a trigonometric substitution. Setting $x = a\sin\theta$ allows us to use standard trigonometric identities to simplify the expressions under the radical signs. Step 1: Apply trigonometric substitution.
Let us substitute $x = \sin\theta$ since the domain is restricted to $x \in (0, 1)$, meaning $\theta \in \left(0, \frac{\pi}{4}\right)$. The differential is: $$dx = \cos\theta\,d\theta$$ Now evaluate the individual radical components inside the integrand using this substitution:
• Bottom radical: $\sqrt{2-x^2} = \sqrt{2-\sin^2\theta}$ Let us try substituting $x = \sqrt{2}\sin\theta$ instead? No, the denominator contains $\sqrt[3]{1-x^2}$, which requires $x = \sin\theta$ or $\cos\theta$ to simplify nicely. Let's look closer at the algebraic identity inside the numerator: Let us analyze the terms inside the roots: $$1 - x\sqrt{2-x^2}$$ If we square the expression $(x + \sqrt{2-x^2})$: $$\left(x + \sqrt{2-x^2}\right)^2 = x^2 + (2-x^2) + 2x\sqrt{2-x^2} = 2 + 2x\sqrt{2-x^2} = 2\left(1 + x\sqrt{2-x^2}\right)$$ This shows that our expressions are algebraically linked! Let's use this relationship to simplify the product in the numerator. Let us instead use the substitution $x = \sin\theta$: $$\sqrt[3]{1-x^2} = \sqrt[3]{\cos^2\theta} = \cos^{\frac{2}{3}}\theta$$ Let's look at another classic trigonometric identity trick for this specific problem type. Let $x = \cos\phi$. An even cleaner approach is to use the substitution $x = \sin\theta$ and analyze the terms together. Let's rewrite the product in the numerator using a common root exponent: Notice that the second term is under a 6th root, while the first term is under a 3rd root. Let's rewrite the 3rd root as a 6th root by squaring its internal expression: $$\sqrt[3]{x+\sqrt{2-x^2}} = \sqrt[6]{\left(x+\sqrt{2-x^2}\right)^2}$$ Using our squared identity from above: $\left(x + \sqrt{2-x^2}\right)^2 = 2\left(1 + x\sqrt{2-x^2}\right)$. Substitute this into the 6th root: $$= \sqrt[6]{2\left(1 + x\sqrt{2-x^2}\right)} = 2^{\frac{1}{6}}\sqrt[6]{1 + x\sqrt{2-x^2}}$$

Step 2: Combine the numerator components.
Now, multiply this term by the second factor in the numerator: $$\text{Numerator} = \left(2^{\frac{1}{6}}\sqrt[6]{1 + x\sqrt{2-x^2}}\right) \times \left(\sqrt[6]{1 - x\sqrt{2-x^2}}\right)$$ Combine the two expressions under a single 6th root using the difference of squares identity $(1+a)(1-a) = 1-a^2$: $$\text{Numerator} = 2^{\frac{1}{6}}\sqrt[6]{\left(1 + x\sqrt{2-x^2}\right)\left(1 - x\sqrt{2-x^2}\right)} = 2^{\frac{1}{6}}\sqrt[6]{1 - x^2(2-x^2)}$$ $$\text{Numerator} = 2^{\frac{1}{6}}\sqrt[6]{1 - 2x^2 + x^4} = 2^{\frac{1}{6}}\sqrt[6]{(1-x^2)^2}$$ Simplify the fractional exponent: $\sqrt[6]{(1-x^2)^2} = (1-x^2)^{\frac{2}{6}} = (1-x^2)^{\frac{1}{3}} = \sqrt[3]{1-x^2}$.

Step 3: Simplify and evaluate the final integral.
Substitute our simplified numerator expression back into the main integral: $$I = \int \frac{2^{\frac{1}{6}}\sqrt[3]{1-x^2}}{\sqrt[3]{1-x^2}}\,dx$$ The complicated radical functions in the numerator and denominator cancel out perfectly: $$I = \int 2^{\frac{1}{6}}\,dx = 2^{\frac{1}{6}}x + C$$ If $2^{1/6}$ can be written in another radical form or if it matches option (A) via a typo or scaling factor. Looking at our choices: (A) $2^{1/12}x+c$, (B) $2^{3/4}x+c$, (C) $2^{1/3}x+c$, (D) $2^{1/6}x+c$. Our derived expression matches choice (D) perfectly! % Final Correct Answer Selection Update Correct Answer: (D) $2^{\frac{1}{6}}x+c$