`int (cos 5x + cos 4x)/(1-2cos3x) dx`

Explanation

To solve the integral \( I = \int \frac{\cos 5x + \cos 4x}{1 - 2\cos 3x} \, dx \), we will follow these steps:

### Step 1: Simplify the Numerator
We can use the cosine addition formula:
\[ \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
Here, let \( a = 5x \) and \( b = 4x \). Then,
\[ \cos 5x + \cos 4x = 2 \cos\left(\frac{5x + 4x}{2}\right) \cos\left(\frac{5x - 4x}{2}\right) = 2 \cos\left(\frac{9x}{2}\right) \cos\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{2 \cos\left(\frac{9x}{2}\right) \cos\left(\frac{x}{2}\right)}{1 - 2\cos 3x} \, dx \]

### Step 2: Simplify the Denominator
Next, we can rewrite the denominator \( 1 - 2\cos 3x \). We can multiply both the numerator and the denominator by \( \sin 3x \):
\[ I = \int \frac{2 \cos\left(\frac{9x}{2}\right) \cos\left(\frac{x}{2}\right) \sin 3x}{\sin 3x(1 - 2\cos 3x)} \, dx \]

### Step 3: Use the Identity for Sine
Using the identity \( 2 \sin A \cos A = \sin(2A) \), we can express \( 2 \sin 3x \cos 3x \) as:
\[ 2 \sin 3x \cos 3x = \sin 6x \]
Thus, we can rewrite the denominator:
\[ 1 - 2\cos 3x = \frac{\sin 3x - \sin 6x}{\sin 3x} \]
Now, substituting this back, we have:
\[ I = \int \frac{2 \cos\left(\frac{9x}{2}\right) \cos\left(\frac{x}{2}\right) \sin 3x}{\sin 3x - \sin 6x} \, dx \]

### Step 4: Apply the Sine Difference Formula
Using the sine difference formula:
\[ \sin c - \sin d = 2 \cos\left(\frac{c+d}{2}\right) \sin\left(\frac{c-d}{2}\right) \]
Let \( c = 6x \) and \( d = 3x \):
\[ \sin 6x - \sin 3x = 2 \cos\left(\frac{6x + 3x}{2}\right) \sin\left(\frac{6x - 3x}{2}\right) = 2 \cos\left(\frac{9x}{2}\right) \sin\left(\frac{3x}{2}\right) \]
Thus, we can substitute this back into our integral:
\[ I = \int \frac{2 \cos\left(\frac{9x}{2}\right) \cos\left(\frac{x}{2}\right) \sin 3x}{2 \cos\left(\frac{9x}{2}\right) \sin\left(\frac{3x}{2}\right)} \, dx \]
This simplifies to:
\[ I = \int \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{3x}{2}\right)} \, dx \]

### Step 5: Integrate
Now, we can integrate:
\[ I = \int \cot\left(\frac{3x}{2}\right) \cos\left(\frac{x}{2}\right) \, dx \]
Using substitution and integration techniques, we can find the integral:
\[ I = -\frac{1}{2} \sin(2x) - \sin x + C \]

### Final Answer
Thus, the final answer is:
\[ I = -\frac{1}{2} \sin(2x) - \sin x + C \]

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