Question

Evaluate: \[ \int_0^a \frac{x^2}{x^6 + a^6} \, dx \]

Hint:

For integrals involving terms like \( x^6 + a^6 \), substitution \( u = x^3 \) often simplifies the problem. Remember to use the standard formula \( \frac{1}{u^2 + b^2} = \frac{1}{b} \tan^{-1} \left( \frac{u}{b} \right) \) for efficient calculation.

Answer 1

This integral can be evaluated using a substitution method. Step 1: Apply substitution. Let \( u = x^3 \), which gives \( du = 3x^2 \, dx \). Adjust the limits of integration: When \( x = 0 \), \( u = 0 \); and when \( x = a \), \( u = a^3 \). The integral becomes: \[ \int_0^{a^3} \frac{1}{u^2 + a^6} \, du. \] Step 2: Use the standard integral formula. For the integral \( \frac{1}{u^2 + b^2} \), the result is: \[ \int \frac{1}{u^2 + b^2} \, du = \frac{1}{b} \tan^{-1} \left( \frac{u}{b} \right). \] Here, \( b = a^3 \), so: \[ \int_0^{a^3} \frac{1}{u^2 + a^6} \, du = \frac{1}{a^3} \tan^{-1} \left( \frac{u}{a^3} \right) \Bigg|_0^{a^3}. \] Step 3: Evaluate the limits. Substitute the limits into the expression: \[ \frac{1}{a^3} \left[ \tan^{-1} \left( \frac{a^3}{a^3} \right) - \tan^{-1} \left( 0 \right) \right]. \] Step 4: Simplify. Simplify the result: \[ = \frac{1}{a^3} \left[ \tan^{-1}(1) - 0 \right] = \frac{1}{a^3} \cdot \frac{\pi}{4}. \] Final Answer: \[ \boxed{\frac{\pi}{4a^3}}. \]