Use substitution and partial fractions to find the indefinite integral. \[ \int \frac{\sin (x)}{\cos (x)+\cos ^{2}(x)} d x \]

To solve the integral \[ \int \frac{\sin(x)}{\cos(x) + \cos^2(x)} \, dx, \] we can begin by using substitution. Let \( u = \cos(x) \). Then, \( du = -\sin(x) \, dx \) or \( -du = \sin(x) \, dx \). The integral becomes: \[ \int \frac{-1}{u + u^2} \, du = -\int \frac{1}{u(1 + u)} \, du. \] Next, we can use partial fraction decomposition on the integrand: \[ \frac{1}{u(1 + u)} = \frac{A}{u} + \frac{B}{1 + u}. \] Multiplying through by the common denominator \( u(1 + u) \) gives: \[ 1 = A(1 + u) + Bu. \] Expanding this, we get: \[ 1 = A + (A + B)u. \] Setting the coefficients equal, we find: 1. For the constant term: \( A = 1 \) 2. For the coefficient of \( u \): \( A + B = 0 \) implies \( B = -1 \) Thus, our partial fractions are: \[ \frac{1}{u(1 + u)} = \frac{1}{u} - \frac{1}{1 + u}. \] Now we can rewrite the integral: \[ -\int \left( \frac{1}{u} - \frac{1}{1 + u} \right) du = -\left( \ln|u| - \ln|1 + u| \right) + C = -\ln\left|\frac{u}{1 + u}\right| + C. \] Finally, substituting back \( u = \cos(x) \): \[ -\ln\left|\frac{\cos(x)}{1 + \cos(x)}\right| + C. \] Thus, our final answer is: \[ -\ln\left|\frac{\cos(x)}{1 + \cos(x)}\right| + C. \]