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

Answer

To solve the integral \( \int \sqrt{2 - 3x^2} \, dx \), we can use a trigonometric substitution. Here’s a step-by-step solution:

### Step 1: Trigonometric Substitution
We notice that the expression under the square root resembles the form \( a^2 - x^2 \). We can rewrite \( 2 - 3x^2 \) as \( 2(1 - \frac{3}{2}x^2) \). To make the substitution easier, we let:
\[ x = \sqrt{\frac{2}{3}} \sin \theta \]
Then, the differential \( dx \) becomes:
\[ dx = \sqrt{\frac{2}{3}} \cos \theta \, d\theta \]

### Step 2: Substitute in the Integral
Substituting \( x \) and \( dx \) into the integral gives:
\[ \int \sqrt{2 - 3\left(\frac{2}{3}\sin^2 \theta\right)} \cdot \sqrt{\frac{2}{3}} \cos \theta \, d\theta \]
This simplifies to:
\[ \int \sqrt{2 - 2\sin^2 \theta} \cdot \sqrt{\frac{2}{3}} \cos \theta \, d\theta \]
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \), we have:
\[ \sqrt{2(1 - \sin^2 \theta)} = \sqrt{2} \cos \theta \]
Thus, the integral becomes:
\[ \int \sqrt{2} \cos \theta \cdot \sqrt{\frac{2}{3}} \cos \theta \, d\theta = \frac{2}{\sqrt{3}} \int \cos^2 \theta \, d\theta \]

### Step 3: Integrate \( \cos^2 \theta \)
To integrate \( \cos^2 \theta \), we use the identity:
\[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \]
Thus, we have:
\[ \int \cos^2 \theta \, d\theta = \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C \]

### Step 4: Substitute Back
Substituting back into our integral:
\[ \frac{2}{\sqrt{3}} \left( \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) \right) + C = \frac{1}{\sqrt{3}} \theta + \frac{1}{2\sqrt{3}} \sin(2\theta) + C \]

### Step 5: Convert Back to \( x \)
Now we need to convert \( \theta \) back to \( x \). From our substitution, we have:
\[ \sin \theta = \frac{\sqrt{3}}{\sqrt{2}} x \implies \theta = \arcsin\left(\frac{\sqrt{3}}{\sqrt{2}} x\right) \]
Also, using the double angle formula:
\[ \sin(2\theta) = 2 \sin \theta \cos \theta = 2 \left(\frac{\sqrt{3}}{\sqrt{2}} x\right) \sqrt{1 - \left(\frac{3}{2} x^2\right)} = \frac{2\sqrt{3}}{\sqrt{2}} x \sqrt{1 - \frac{3}{2} x^2} \]

### Final Result
Thus, the final answer is:
\[ \int \sqrt{2 - 3x^2} \, dx = \frac{1}{\sqrt{3}} \arcsin\left(\frac{\sqrt{3}}{\sqrt{2}} x\right) + \frac{1}{2\sqrt{3}} \left( \frac{2\sqrt{3}}{\sqrt{2}} x \sqrt{1 - \frac{3}{2} x^2} \right) + C \]