Evaluate : `int(dx)/(sqrt(5-4x^(2))`

Answer

To evaluate the integral \( I = \int \frac{dx}{\sqrt{5 - 4x^2}} \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{5 - 4x^2}} \]
To simplify this, we can factor out the constant from the square root.

### Step 2: Factor out the constant
We can rewrite the expression under the square root:
\[ I = \int \frac{dx}{\sqrt{5(1 - \frac{4}{5}x^2)}} \]
This simplifies to:
\[ I = \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{1 - \frac{4}{5}x^2}} \]

### Step 3: Use a trigonometric substitution
We can use the substitution \( x = \frac{\sqrt{5}}{2} \sin \theta \). Then, we differentiate to find \( dx \):
\[ dx = \frac{\sqrt{5}}{2} \cos \theta \, d\theta \]
Now, we substitute \( x \) and \( dx \) into the integral:
\[ I = \frac{1}{\sqrt{5}} \int \frac{\frac{\sqrt{5}}{2} \cos \theta \, d\theta}{\sqrt{1 - \frac{4}{5} \left(\frac{\sqrt{5}}{2} \sin \theta\right)^2}} \]

### Step 4: Simplify the integral
Now, we simplify the expression inside the square root:
\[ 1 - \frac{4}{5} \left(\frac{5}{4} \sin^2 \theta\right) = 1 - \sin^2 \theta = \cos^2 \theta \]
Thus, the integral becomes:
\[ I = \frac{1}{\sqrt{5}} \int \frac{\frac{\sqrt{5}}{2} \cos \theta \, d\theta}{\cos \theta} \]
This simplifies to:
\[ I = \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{2} \int d\theta = \frac{1}{2} \int d\theta \]

### Step 5: Integrate
Integrating \( d\theta \) gives:
\[ I = \frac{1}{2} \theta + C \]

### Step 6: Substitute back for \( \theta \)
Recall the substitution \( x = \frac{\sqrt{5}}{2} \sin \theta \), which implies:
\[ \sin \theta = \frac{2x}{\sqrt{5}} \quad \Rightarrow \quad \theta = \sin^{-1}\left(\frac{2x}{\sqrt{5}}\right) \]
Thus, we have:
\[ I = \frac{1}{2} \sin^{-1}\left(\frac{2x}{\sqrt{5}}\right) + C \]

### Final Answer
The final result for the integral is:
\[ I = \frac{1}{2} \sin^{-1}\left(\frac{2x}{\sqrt{5}}\right) + C \]