integrate: `int(sin6x+sin4x)/(cos6x+cos4x)dx`

Answer

To solve the integral \( \int \frac{\sin 6x + \sin 4x}{\cos 6x + \cos 4x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Sine and Cosine Functions
We can use the sum-to-product identities for sine and cosine:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

For \( A = 6x \) and \( B = 4x \):
- \( A + B = 10x \)
- \( A - B = 2x \)

Thus, we can rewrite:
\[ \sin 6x + \sin 4x = 2 \sin(5x) \cos(x) \]
\[ \cos 6x + \cos 4x = 2 \cos(5x) \cos(x) \]

### Step 2: Substitute into the Integral
Substituting these into the integral gives:
\[ \int \frac{2 \sin(5x) \cos(x)}{2 \cos(5x) \cos(x)} \, dx \]
The \( 2 \) and \( \cos(x) \) terms cancel out:
\[ \int \frac{\sin(5x)}{\cos(5x)} \, dx \]

### Step 3: Simplify the Integral
The expression simplifies to:
\[ \int \tan(5x) \, dx \]

### Step 4: Integrate
The integral of \( \tan(kx) \) is given by:
\[ \int \tan(kx) \, dx = -\frac{1}{k} \ln |\cos(kx)| + C \]
In our case, \( k = 5 \):
\[ \int \tan(5x) \, dx = -\frac{1}{5} \ln |\cos(5x)| + C \]

### Final Answer
Thus, the final answer is:
\[ -\frac{1}{5} \ln |\cos(5x)| + C \]

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