Evaluate `int (dx)/((5+4cos x)^(2))`.

Answer

To evaluate the integral

\[ I = \int \frac{dx}{(5 + 4 \cos x)^2} \]

we can use a substitution method and some trigonometric identities. Here’s a step-by-step solution:

### Step 1: Use the substitution
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, 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: Substitute into the integral
Substituting these into the integral, we get:

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

### Step 3: Simplify the expression
Now simplify the expression inside the integral:

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

Thus,

\[ (5 + 4 \cos x)^2 = \left(\frac{9 - t^2}{1 + t^2}\right)^2 = \frac{(9 - t^2)^2}{(1 + t^2)^2}. \]

### Step 4: Substitute back into the integral
Now substituting this back into the integral, we have:

\[ I = \int \frac{2(1 + t^2)}{(9 - t^2)^2} dt. \]

### Step 5: Split the integral
This can be split into two parts:

\[ I = 2 \int \frac{1}{(9 - t^2)^2} dt + 2 \int \frac{t^2}{(9 - t^2)^2} dt. \]

### Step 6: Evaluate the first integral
The first integral can be solved using the formula:

\[ \int \frac{1}{(a^2 - x^2)^2} dx = \frac{x}{2a^2(a^2 - x^2)} + \frac{1}{2a^4} \tan^{-1}\left(\frac{x}{a}\right) + C. \]

Here, \( a = 3 \), so:

\[ \int \frac{1}{(9 - t^2)^2} dt = \frac{t}{18(9 - t^2)} + \frac{1}{18} \tan^{-1}\left(\frac{t}{3}\right) + C. \]

### Step 7: Evaluate the second integral
The second integral can be simplified using the substitution \( u = 9 - t^2 \):

\[ \int \frac{t^2}{(9 - t^2)^2} dt = -\frac{1}{9} \int \frac{du}{u^2} = \frac{1}{9u} + C = \frac{1}{9(9 - t^2)} + C. \]

### Step 8: Combine results
Now combine the results of both integrals to find \( I \):

\[ I = 2\left[\frac{t}{18(9 - t^2)} + \frac{1}{18} \tan^{-1}\left(\frac{t}{3}\right) + \frac{1}{9(9 - t^2)}\right]. \]

### Step 9: Substitute back for \( t \)
Finally, substitute back \( t = \tan\left(\frac{x}{2}\right) \) to express \( I \) in terms of \( x \):

\[ I = \frac{1}{9(9 - \tan^2\left(\frac{x}{2}\right))} + \frac{1}{9(9 - \tan^2\left(\frac{x}{2}\right))} + \frac{1}{9} \tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right). \]

### Final Result
Thus, the evaluated integral is:

\[ I = \frac{2}{9(9 - \tan^2\left(\frac{x}{2}\right))} + \frac{1}{9} \tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right) + C. \]