`int_(0)^(pi//2)(1)/(5+4cos x)dx=`

Options

A. `(2)/(3)tan^(-1)((1)/(3))`
B. `(pi)/(3)`
C. `(1)/(3)tan^(-1)((2)/(3))`
D. `(pi)/(6)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{1}{5 + 4 \cos x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral using a trigonometric substitution
We can use the substitution \( \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \). This leads us to express \( \cos x \) in terms of \( t = \tan\left(\frac{x}{2}\right) \).

### Step 2: Change the limits of integration
When \( x = 0 \), \( t = \tan(0) = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \).

### Step 3: Substitute \( \cos x \) and \( dx \)
Using the substitution, we have:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \]
The differential \( dx \) can be expressed as:
\[ dx = \frac{2}{1 + t^2} \, dt \]

### Step 4: Substitute into the integral
Now, substituting \( \cos x \) and \( dx \) into the integral:
\[ I = \int_{0}^{1} \frac{2}{5 + 4 \left(\frac{1 - t^2}{1 + t^2}\right)} \cdot \frac{1}{1 + t^2} \, dt \]
This simplifies to:
\[ I = \int_{0}^{1} \frac{2(1 + t^2)}{5(1 + t^2) + 4(1 - t^2)} \, dt \]
\[ = \int_{0}^{1} \frac{2(1 + t^2)}{9 + t^2} \, dt \]

### Step 5: Simplify the integral
Now, we can separate the integral:
\[ I = \int_{0}^{1} \frac{2}{9 + t^2} \, dt + \int_{0}^{1} \frac{2t^2}{9 + t^2} \, dt \]

### Step 6: Solve the first integral
The first integral can be solved using the formula for arctangent:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \]
Thus,
\[ \int_{0}^{1} \frac{2}{9 + t^2} \, dt = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 7: Solve the second integral
For the second integral, we can use integration by parts or another substitution. However, we can also note that:
\[ \int_{0}^{1} \frac{2t^2}{9 + t^2} \, dt = \int_{0}^{1} \left( \frac{2}{9 + t^2} - \frac{18}{9 + t^2} \right) dt \]
This simplifies to:
\[ = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right) - 6 \int_{0}^{1} \frac{1}{9 + t^2} \, dt \]

### Step 8: Combine results
Combining the results from both integrals, we find:
\[ I = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right) \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right) \]