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

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{5 + 4 \cos x} \, dx, \]

we can use a substitution involving the tangent function. Let's proceed step by step.

### Step 1: Set up the integral
We start with the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{5 + 4 \cos x} \, dx. \]

### Step 2: Use the substitution \( t = \tan\left(\frac{x}{2}\right) \)
Using the Weierstrass substitution, we have:

\[ \cos x = \frac{1 - t^2}{1 + t^2} \quad \text{and} \quad dx = \frac{2}{1 + t^2} \, dt. \]

When \( x = 0 \), \( t = 0 \), and when \( x = \frac{\pi}{2} \), \( t \to 1 \).

### Step 3: Substitute into the integral
Substituting these into the integral, we get:

\[ I = \int_{0}^{1} \frac{2}{1 + t^2} \cdot \frac{1}{5 + 4 \cdot \frac{1 - t^2}{1 + t^2}} \, dt. \]

### Step 4: Simplify the integrand
Now simplify the integrand:

\[ 5 + 4 \cdot \frac{1 - t^2}{1 + t^2} = 5 + \frac{4(1 - t^2)}{1 + t^2} = \frac{5(1 + t^2) + 4(1 - t^2)}{1 + t^2} = \frac{5 + 5t^2 + 4 - 4t^2}{1 + t^2} = \frac{9 + t^2}{1 + t^2}. \]

Thus, the integral becomes:

\[ I = \int_{0}^{1} \frac{2}{1 + t^2} \cdot \frac{1 + t^2}{9 + t^2} \, dt = \int_{0}^{1} \frac{2}{9 + t^2} \, dt. \]

### Step 5: Evaluate the integral
Now we evaluate:

\[ I = 2 \int_{0}^{1} \frac{1}{9 + t^2} \, dt. \]

The integral \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \). Here, \( a^2 = 9 \) so \( a = 3 \):

\[ \int \frac{1}{9 + t^2} \, dt = \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right). \]

Thus, we have:

\[ I = 2 \left[ \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right) \right]_{0}^{1} = \frac{2}{3} \left( \tan^{-1}\left(\frac{1}{3}\right) - \tan^{-1}(0) \right) = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right). \]

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

\[ I = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right). \]

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