Evaluate the following integrals:
`int(dx)/((3cosx+4sinx))`

Explanation

To evaluate the integral

\[ I = \int \frac{dx}{3 \cos x + 4 \sin x} \]

we can follow these steps:

### Step 1: Rewrite the Denominator
We start by rewriting the denominator \(3 \cos x + 4 \sin x\) in a more manageable form. We can express it in terms of a single trigonometric function using the amplitude-phase form:

\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]

Now, we can find the angle \(\alpha\) such that:

\[ \cos \alpha = \frac{3}{5}, \quad \sin \alpha = \frac{4}{5} \]

Thus, we can rewrite the denominator as:

\[ 3 \cos x + 4 \sin x = 5 \left( \frac{3}{5} \cos x + \frac{4}{5} \sin x \right) = 5 \cos(x - \alpha) \]

### Step 2: Substitute in the Integral
Now we substitute this back into the integral:

\[ I = \int \frac{dx}{5 \cos(x - \alpha)} = \frac{1}{5} \int \sec(x - \alpha) \, dx \]

### Step 3: Integrate the Secant Function
The integral of \(\sec u\) is known:

\[ \int \sec u \, du = \ln | \sec u + \tan u | + C \]

So we have:

\[ I = \frac{1}{5} \left( \ln | \sec(x - \alpha) + \tan(x - \alpha) | + C \right) \]

### Step 4: Substitute Back for \(\alpha\)
Now we substitute back for \(\alpha\):

\[ I = \frac{1}{5} \left( \ln | \sec(x - \alpha) + \tan(x - \alpha) | + C \right) \]

### Step 5: Express in Terms of Original Variables
Using the definitions of \(\sec\) and \(\tan\):

\[ \sec(x - \alpha) = \frac{1}{\cos(x - \alpha)}, \quad \tan(x - \alpha) = \frac{\sin(x - \alpha)}{\cos(x - \alpha)} \]

Thus, we can express the final result as:

\[ I = \frac{1}{5} \ln \left| \frac{1 + \tan(x - \alpha)}{\cos(x - \alpha)} \right| + C \]

### Final Result
The final result of the integral is:

\[ I = \frac{1}{5} \ln \left| \frac{5 \sin x + 3 \cos x}{5} \right| + C \]