`int(dx)/(sqrt(4-9x^(2)))=?`

Options

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

Explanation

To solve the integral \( \int \frac{dx}{\sqrt{4 - 9x^2}} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{4 - 9x^2}} \]

### Step 2: Factor Out Constants
Notice that \(4\) can be expressed as \((\frac{2}{3})^2\) multiplied by \(9\). To facilitate integration, we rewrite the expression under the square root:
\[ I = \int \frac{dx}{\sqrt{4(1 - \frac{9}{4}x^2)}} = \int \frac{dx}{\sqrt{4} \sqrt{1 - \left(\frac{3}{2}x\right)^2}} \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{dx}{\sqrt{1 - \left(\frac{3}{2}x\right)^2}} \]

### Step 3: Use Substitution
Let \(u = \frac{3}{2}x\), then \(dx = \frac{2}{3} du\). Substituting this into the integral gives:
\[ I = \frac{1}{2} \int \frac{\frac{2}{3} du}{\sqrt{1 - u^2}} = \frac{1}{3} \int \frac{du}{\sqrt{1 - u^2}} \]

### Step 4: Integrate
The integral \( \int \frac{du}{\sqrt{1 - u^2}} \) is a standard integral that equals \( \sin^{-1}(u) + C \):
\[ I = \frac{1}{3} \sin^{-1}(u) + C \]

### Step 5: Substitute Back
Now, substitute back \(u = \frac{3}{2}x\):
\[ I = \frac{1}{3} \sin^{-1}\left(\frac{3}{2}x\right) + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{dx}{\sqrt{4 - 9x^2}} = \frac{1}{3} \sin^{-1}\left(\frac{3}{2}x\right) + C \]

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