`int sqrt(2-3x^(2))dx`

Answer

To solve the integral \( \int \sqrt{2 - 3x^2} \, dx \), we will follow a series of steps to simplify and evaluate the integral.

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \sqrt{2 - 3x^2} \, dx \]
We can factor out the constant from the square root:
\[ = \int \sqrt{3} \sqrt{\frac{2}{3} - x^2} \, dx \]
This allows us to take \( \sqrt{3} \) outside of the integral:
\[ = \sqrt{3} \int \sqrt{\frac{2}{3} - x^2} \, dx \]

### Step 2: Use a Trigonometric Substitution
We can use the identity for the integral of the form \( \int \sqrt{a^2 - x^2} \, dx \). Here, we identify \( a^2 = \frac{2}{3} \), so \( a = \sqrt{\frac{2}{3}} \).

Using the substitution \( x = a \sin(\theta) \), we have:
\[ dx = a \cos(\theta) \, d\theta \]
Substituting \( x = \sqrt{\frac{2}{3}} \sin(\theta) \) into the integral gives:
\[ = \sqrt{3} \int \sqrt{\frac{2}{3} - \left(\sqrt{\frac{2}{3}} \sin(\theta)\right)^2} \cdot \sqrt{\frac{2}{3}} \cos(\theta) \, d\theta \]
This simplifies to:
\[ = \sqrt{3} \int \sqrt{\frac{2}{3} (1 - \sin^2(\theta))} \cdot \sqrt{\frac{2}{3}} \cos(\theta) \, d\theta \]
Since \( 1 - \sin^2(\theta) = \cos^2(\theta) \), we have:
\[ = \sqrt{3} \int \sqrt{\frac{2}{3}} \cos(\theta) \cdot \sqrt{\frac{2}{3}} \cos(\theta) \, d\theta \]
\[ = \frac{2}{3} \sqrt{3} \int \cos^2(\theta) \, d\theta \]

### Step 3: Integrate \( \cos^2(\theta) \)
Using the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \):
\[ = \frac{2}{3} \sqrt{3} \int \frac{1 + \cos(2\theta)}{2} \, d\theta \]
\[ = \frac{1}{3} \sqrt{3} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C \]

### Step 4: Back Substitute
Now we need to back substitute \( \theta \) in terms of \( x \). Recall that:
\[ \sin(\theta) = \frac{\sqrt{3}}{\sqrt{2}} x \]
Thus:
\[ \theta = \sin^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}} x\right) \]
Substituting back gives:
\[ = \frac{1}{3} \sqrt{3} \left( \sin^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}} x\right) + \frac{1}{2} \sin\left(2 \sin^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}} x\right)\right) \right) + C \]

### Final Answer
The final answer for the integral is:
\[ \int \sqrt{2 - 3x^2} \, dx = \frac{1}{3} \sqrt{3} \left( \sin^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}} x\right) + \frac{1}{2} \sin\left(2 \sin^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}} x\right)\right) \right) + C \]