\( \int \frac { d x } { 3 \sin x + 4 \cos x } \)

To solve the integral \[ I = \int \frac{dx}{3 \sin x + 4 \cos x}, \] we can use a substitution to simplify the expression in the denominator. We can express \(3 \sin x + 4 \cos x\) in a different form. First, we can rewrite \(3 \sin x + 4 \cos x\) as \(R \sin(x + \phi)\), where \(R\) and \(\phi\) are determined as follows: 1. Calculate \(R\): \[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \] 2. Determine \(\phi\) such that: \[ \cos \phi = \frac{4}{5}, \quad \sin \phi = \frac{3}{5}. \] Thus, we can rewrite the integral as: \[ I = \int \frac{dx}{5 \sin(x + \phi)} = \frac{1}{5} \int \frac{dx}{\sin(x + \phi)}. \] The integral of \(\frac{1}{\sin u}\) is known: \[ \int \frac{du}{\sin u} = \ln \left| \tan \frac{u}{2} \right| + C. \] Substituting back, we have: \[ I = \frac{1}{5} \ln \left| \tan \frac{x + \phi}{2} \right| + C. \] Now, substituting \(\phi\) back in terms of \(\sin\) and \(\cos\): \[ \phi = \tan^{-1}\left(\frac{3}{4}\right). \] Thus, the final result for the integral is: \[ \int \frac{dx}{3 \sin x + 4 \cos x} = \frac{1}{5} \ln \left| \tan \left( \frac{x + \tan^{-1}\left(\frac{3}{4}\right)}{2} \right) \right| + C. \]