Evaluate : `int_(0)^((pi)/(4)) (cos x)/(4-sin^(2)x)dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{4 - \sin^2 x} \, dx, \]

we will use a substitution method. Let's follow the steps:

### Step 1: Substitution
Let \( t = \sin x \). Then, the derivative \( dt = \cos x \, dx \).

When \( x = 0 \), \( t = \sin(0) = 0 \).
When \( x = \frac{\pi}{4} \), \( t = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).

Thus, we can rewrite the integral in terms of \( t \):

\[ I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{4 - t^2} \, dt. \]

### Step 2: Simplifying the Integral
The integral can be expressed as:

\[ I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{4 - t^2} \, dt. \]

We can recognize that \( 4 - t^2 \) can be factored as \( 2^2 - t^2 \). This suggests the use of the formula for the integral of the form \( \int \frac{1}{a^2 - x^2} \, dx \).

### Step 3: Using the Integral Formula
The formula for the integral is:

\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \log\left|\frac{a+x}{a-x}\right| + C. \]

In our case, \( a = 2 \). Thus, we have:

\[ I = \frac{1}{2 \cdot 2} \log\left|\frac{2 + t}{2 - t}\right| \bigg|_0^{\frac{1}{\sqrt{2}}}. \]

### Step 4: Evaluating the Limits
Now we evaluate the limits:

1. When \( t = \frac{1}{\sqrt{2}} \):

\[ \frac{2 + \frac{1}{\sqrt{2}}}{2 - \frac{1}{\sqrt{2}}} = \frac{2 + \frac{1}{\sqrt{2}}}{2 - \frac{1}{\sqrt{2}}} = \frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}. \]

2. When \( t = 0 \):

\[ \frac{2 + 0}{2 - 0} = 1. \]

Thus, we have:

\[ I = \frac{1}{4} \left( \log\left(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\right) - \log(1) \right). \]

Since \( \log(1) = 0 \), we simplify to:

\[ I = \frac{1}{4} \log\left(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\right). \]

### Final Result
Thus, the evaluated integral is:

\[ I = \frac{1}{4} \log\left(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}\right). \]