`int sqrt(1-9x^(2))dx=?`

Options

A. `(x)/(2)sqrt(1-9x^(2))+(1)/(18)sin^(-1)3x+C`
B. `(3x)/(2)sqrt(1-9x^(2))+(1)/(6)sin^(-1)3x+C`
C. `(x)/(2)sqrt(1-9x^(2))+(1)/(6)sin^(-1)3x+C`
D. None of these

Explanation

To solve the integral \( \int \sqrt{1 - 9x^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \sqrt{1 - 9x^2} \, dx \]

### Step 2: Factor out the coefficient of \(x^2\)
Notice that \(9x^2\) can be rewritten as \((3x)^2\). Thus, we can rewrite the integral as:
\[ I = \int \sqrt{1 - (3x)^2} \, dx \]

### Step 3: Use a Trigonometric Substitution
We can use the substitution \(3x = \sin(\theta)\), which implies:
\[ x = \frac{1}{3} \sin(\theta) \quad \text{and} \quad dx = \frac{1}{3} \cos(\theta) \, d\theta \]
Substituting these into the integral gives:
\[ I = \int \sqrt{1 - \sin^2(\theta)} \cdot \frac{1}{3} \cos(\theta) \, d\theta \]

### Step 4: Simplify the Integral
Using the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\), we can simplify:
\[ I = \int \cos(\theta) \cdot \frac{1}{3} \cos(\theta) \, d\theta = \frac{1}{3} \int \cos^2(\theta) \, d\theta \]

### Step 5: Use the Power Reduction Formula
We can use the power reduction formula:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Thus, the integral becomes:
\[ I = \frac{1}{3} \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{6} \int (1 + \cos(2\theta)) \, d\theta \]

### Step 6: Integrate
Now we can integrate:
\[ I = \frac{1}{6} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C \]

### Step 7: Substitute Back
Recall that \(3x = \sin(\theta)\), so \(\theta = \sin^{-1}(3x)\). Therefore, we can substitute back:
\[ I = \frac{1}{6} \left( \sin^{-1}(3x) + \frac{1}{2} \sin(2 \sin^{-1}(3x)) \right) + C \]

### Step 8: Simplify Further
Using the double angle identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):
\[ \sin(2 \sin^{-1}(3x)) = 2(3x)\sqrt{1 - (3x)^2} = 6x\sqrt{1 - 9x^2} \]
Thus, we have:
\[ I = \frac{1}{6} \left( \sin^{-1}(3x) + 3x\sqrt{1 - 9x^2} \right) + C \]

### Final Result
The final result for the integral is:
\[ I = \frac{1}{6} \sin^{-1}(3x) + \frac{1}{2} x \sqrt{1 - 9x^2} + C \]