`int_(0)^(pi//2) (1)/(4+3 cos x)dx`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 3 \cos x} \, dx, \]

we can use a substitution involving the tangent function. Here’s a step-by-step solution:

### Step 1: Use the substitution \( t = \tan\left(\frac{x}{2}\right) \)

Using the Weierstrass substitution, we have:

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

### Step 2: Change the limits of integration

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

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

### Step 3: Substitute into the integral

Now substituting into the integral:

\[ I = \int_{0}^{1} \frac{1}{4 + 3 \left(\frac{1 - t^2}{1 + t^2}\right)} \cdot \frac{2}{1 + t^2} \, dt. \]

### Step 4: Simplify the integrand

The integrand becomes:

\[ I = \int_{0}^{1} \frac{2}{(4 + 3 \frac{1 - t^2}{1 + t^2})(1 + t^2)} \, dt. \]

Now simplifying the term in the denominator:

\[ 4 + 3 \frac{1 - t^2}{1 + t^2} = \frac{(4 + 3)(1 + t^2) + 3(1 - t^2)}{1 + t^2} = \frac{7 + 4t^2}{1 + t^2}. \]

Thus, we have:

\[ I = \int_{0}^{1} \frac{2(1 + t^2)}{7 + 4t^2} \, dt. \]

### Step 5: Split the integral

This can be split into two parts:

\[ I = 2 \int_{0}^{1} \frac{1}{7 + 4t^2} \, dt + 2 \int_{0}^{1} \frac{t^2}{7 + 4t^2} \, dt. \]

### Step 6: Evaluate the first integral

The first integral can be evaluated using the formula:

\[ \int \frac{1}{a + bx^2} \, dx = \frac{1}{\sqrt{ab}} \tan^{-1}\left(\frac{\sqrt{b}}{\sqrt{a}} x\right) + C. \]

For \( a = 7 \) and \( b = 4 \):

\[ \int_{0}^{1} \frac{1}{7 + 4t^2} \, dt = \frac{1}{\sqrt{7 \cdot 4}} \tan^{-1}\left(\frac{2}{\sqrt{7}} t\right) \bigg|_0^1 = \frac{1}{2\sqrt{7}} \left(\tan^{-1}\left(\frac{2}{\sqrt{7}}\right) - \tan^{-1}(0)\right). \]

Thus,

\[ \int_{0}^{1} \frac{1}{7 + 4t^2} \, dt = \frac{1}{2\sqrt{7}} \tan^{-1}\left(\frac{2}{\sqrt{7}}\right). \]

### Step 7: Evaluate the second integral

For the second integral, we can use integration by parts or another substitution. However, we can also use the result from the first integral to find:

\[ \int_{0}^{1} \frac{t^2}{7 + 4t^2} \, dt = \frac{1}{4} \int_{0}^{1} \frac{1}{7 + 4t^2} \, dt - \frac{1}{4} \int_{0}^{1} \frac{1}{7 + 4t^2} \, dt. \]

### Final Result

Combining both integrals and simplifying gives us the final result:

\[ I = \frac{1}{2\sqrt{7}} \tan^{-1}\left(\frac{2}{\sqrt{7}}\right). \]