`int(1)/(7+5 cos x)dx=`

Options

A. `(1)/(sqrt(6))tan^(-1)((1)/(sqrt(6))"tan" (x)/(2))+C`.
B. `(1)/(sqrt(3))tan^(-1)((1)/(sqrt(3))"tan" (x)/(2))+C`
C. `(1)/(4)tan^(-1)((x)/(2))+C`
D. `(1)/(7)tan^(-1)("tan" (x)/(2))+C`

Explanation

To solve the integral \( I = \int \frac{1}{7 + 5 \cos x} \, dx \), we will use a trigonometric substitution and some algebraic manipulation. Here’s a step-by-step solution:

### Step 1: Rewrite the cosine term
We can express \( \cos x \) in terms of \( \tan \frac{x}{2} \) using the identity:
\[ \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]
Let \( t = \tan \frac{x}{2} \). Then, we have:
\[ \cos x = \frac{1 - t^2}{1 + t^2} \]

### Step 2: Substitute in the integral
Substituting this into the integral gives:
\[ I = \int \frac{1}{7 + 5 \left( \frac{1 - t^2}{1 + t^2} \right)} \, dx \]
This simplifies to:
\[ I = \int \frac{1 + t^2}{7(1 + t^2) + 5(1 - t^2)} \, dx \]
Simplifying the denominator:
\[ 7(1 + t^2) + 5(1 - t^2) = 7 + 7t^2 + 5 - 5t^2 = 12 + 2t^2 \]
Thus, the integral becomes:
\[ I = \int \frac{1 + t^2}{12 + 2t^2} \, dx \]

### Step 3: Factor out constants
Factor out the constant from the denominator:
\[ I = \int \frac{1 + t^2}{2(6 + t^2)} \, dx = \frac{1}{2} \int \frac{1 + t^2}{6 + t^2} \, dx \]

### Step 4: Change of variable for \( dx \)
Now, we need to express \( dx \) in terms of \( dt \):
Using the derivative of \( t = \tan \frac{x}{2} \):
\[ dx = \frac{2}{1 + t^2} \, dt \]
Substituting this into the integral:
\[ I = \frac{1}{2} \int \frac{1 + t^2}{6 + t^2} \cdot \frac{2}{1 + t^2} \, dt = \int \frac{1}{6 + t^2} \, dt \]

### Step 5: Solve the integral
The integral \( \int \frac{1}{6 + t^2} \, dt \) is a standard form:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
Here, \( a^2 = 6 \) so \( a = \sqrt{6} \):
\[ I = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{t}{\sqrt{6}} \right) + C \]

### Step 6: Substitute back for \( t \)
Recall that \( t = \tan \frac{x}{2} \):
\[ I = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{\tan \frac{x}{2}}{\sqrt{6}} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{\tan \frac{x}{2}}{\sqrt{6}} \right) + C \]

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