Evaluate: `intx^2sqrt(a^6-x^6)dx`

Answer

To evaluate the integral

\[ I = \int x^2 \sqrt{a^6 - x^6} \, dx, \]

we will use a substitution method. Let's go through the solution step by step.

### Step 1: Substitution
Let \( t = x^3 \). Then, differentiating both sides gives us:

\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2}. \]

### Step 2: Expressing \( x^2 \, dx \)
From the substitution, we can express \( x^2 \, dx \) as:

\[ x^2 \, dx = \frac{dt}{3}. \]

### Step 3: Rewriting the Integral
Now we can rewrite the integral in terms of \( t \):

\[ I = \int x^2 \sqrt{a^6 - x^6} \, dx = \int \frac{dt}{3} \sqrt{a^6 - t^2}. \]

### Step 4: Factor Out the Constant
We can factor out the constant \( \frac{1}{3} \):

\[ I = \frac{1}{3} \int \sqrt{a^6 - t^2} \, dt. \]

### Step 5: Recognizing the Integral Form
The integral \( \int \sqrt{a^2 - x^2} \, dx \) has a known solution. In our case, we can rewrite \( a^6 \) as \( (a^3)^2 \):

\[ I = \frac{1}{3} \int \sqrt{(a^3)^2 - t^2} \, dt. \]

### Step 6: Using the Integral Formula
Using the formula for the integral \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C \):

Here, \( a = a^3 \) and \( x = t \):

\[ I = \frac{1}{3} \left( \frac{t}{2} \sqrt{(a^3)^2 - t^2} + \frac{(a^3)^2}{2} \sin^{-1} \left( \frac{t}{a^3} \right) \right) + C. \]

### Step 7: Substitute Back \( t = x^3 \)
Now we substitute back \( t = x^3 \):

\[ I = \frac{1}{3} \left( \frac{x^3}{2} \sqrt{a^6 - x^6} + \frac{a^6}{2} \sin^{-1} \left( \frac{x^3}{a^3} \right) \right) + C. \]

### Final Answer
Thus, the final result is:

\[ I = \frac{x^3}{6} \sqrt{a^6 - x^6} + \frac{a^6}{6} \sin^{-1} \left( \frac{x^3}{a^3} \right) + C. \]