`int (x^(2))/(sqrt(1-x^(6)))dx=`….

Options

A. `-sin^(-1) (x^(3))+c`
B. `cos^(-1) (x^(3))+c`
C. `sin^(-1) (x^(3))+c`
D. `(1)/(3) sin^(-1) (x^(3))+c`

Answer

To solve the integral \( \int \frac{x^2}{\sqrt{1 - x^6}} \, dx \), we can use a substitution method. Here are the steps to find the solution:

### Step 1: Substitution
Let \( t = x^3 \). Then, we differentiate to find \( dt \):
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
We can express \( x^2 \) in terms of \( t \):
\[ x^2 = t^{2/3} \]

### Step 2: Rewrite the integral
Substituting \( x^2 \) and \( dx \) into the integral, we have:
\[ \int \frac{x^2}{\sqrt{1 - x^6}} \, dx = \int \frac{t^{2/3}}{\sqrt{1 - t^2}} \cdot \frac{dt}{3t^{2/3}} = \frac{1}{3} \int \frac{1}{\sqrt{1 - t^2}} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{\sqrt{1 - t^2}} \, dt \) is a standard integral, which gives:
\[ \int \frac{1}{\sqrt{1 - t^2}} \, dt = \sin^{-1}(t) + C \]
Thus, we can write:
\[ \frac{1}{3} \int \frac{1}{\sqrt{1 - t^2}} \, dt = \frac{1}{3} \sin^{-1}(t) + C \]

### Step 4: Back-substitute
Now, we substitute back \( t = x^3 \):
\[ \frac{1}{3} \sin^{-1}(t) + C = \frac{1}{3} \sin^{-1}(x^3) + C \]

### Final Answer
Therefore, the solution to the integral is:
\[ \int \frac{x^2}{\sqrt{1 - x^6}} \, dx = \frac{1}{3} \sin^{-1}(x^3) + C \]

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