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

Options

A. `(1)/(4)log 3`
B. `(1)/(3)log 4`
C. `4log((1)/(3))`
D. `(1)/(2)log 3`

Answer

To solve the integral

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

we will perform a substitution and simplify the integral step by step.

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

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

Thus, the limits of integration change from \( x = 0 \) to \( x = \frac{\pi}{2} \) into \( t = 0 \) to \( t = 1 \).

Now, we can rewrite the integral:

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

### Step 2: Simplifying the Integral
We can factor the denominator:

\[ 4 - t^2 = (2 - t)(2 + t). \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{1} \frac{1}{(2 - t)(2 + t)} \, dt. \]

### Step 3: Partial Fraction Decomposition
We can express the integrand using partial fractions:

\[ \frac{1}{(2 - t)(2 + t)} = \frac{A}{2 - t} + \frac{B}{2 + t}. \]

Multiplying through by the denominator \( (2 - t)(2 + t) \):

\[ 1 = A(2 + t) + B(2 - t). \]

Expanding this gives:

\[ 1 = (2A + 2B) + (A - B)t. \]

Setting up the system of equations:

1. \( 2A + 2B = 1 \)
2. \( A - B = 0 \)

From the second equation, we have \( A = B \). Substituting \( B = A \) into the first equation:

\[ 2A + 2A = 1 \implies 4A = 1 \implies A = \frac{1}{4}, \, B = \frac{1}{4}. \]

Thus, we can rewrite the integral:

\[ I = \int_{0}^{1} \left( \frac{1/4}{2 - t} + \frac{1/4}{2 + t} \right) dt. \]

### Step 4: Integrating
Now we can integrate each term separately:

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

The integrals can be computed as follows:

1. For \( \int \frac{1}{2 - t} \, dt = -\log|2 - t| \).
2. For \( \int \frac{1}{2 + t} \, dt = \log|2 + t| \).

Thus, we have:

\[ I = \frac{1}{4} \left[ -\log|2 - t| \right]_{0}^{1} + \frac{1}{4} \left[ \log|2 + t| \right]_{0}^{1}. \]

Calculating these limits:

1. For \( -\log|2 - t| \) from \( 0 \) to \( 1 \):

\[ -\log|2 - 1| + \log|2 - 0| = -\log(1) + \log(2) = 0 + \log(2) = \log(2). \]

2. For \( \log|2 + t| \) from \( 0 \) to \( 1 \):

\[ \log|2 + 1| - \log|2 + 0| = \log(3) - \log(2) = \log\left(\frac{3}{2}\right). \]

Combining these results:

\[ I = \frac{1}{4} \left( \log(2) + \log\left(\frac{3}{2}\right) \right) = \frac{1}{4} \log\left(2 \cdot \frac{3}{2}\right) = \frac{1}{4} \log(3). \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{1}{4} \log(3). \]