11 Points] DETAILS MY NOTES LARCALC12 8.5.025. Use substitution and partial fractions to find the indefinite integral. (Remember to use absolute valiues where aporopriate Remember the constant of integration.) \( \int \frac{\sin (x)}{\cos (x)+\cos ^{2}(x)} d x \)

To solve the integral \( \int \frac{\sin(x)}{\cos(x) + \cos^2(x)} dx \), we start by using a substitution. Set \( u = \cos(x) \), then \( du = -\sin(x) dx \) or \( -du = \sin(x) dx \). The integral now becomes: \[ -\int \frac{1}{u + u^2} du \] Next, we can simplify the integrand using partial fractions. Rewrite \( \frac{1}{u + u^2} \) as \( \frac{1}{u(1+u)} \) and decompose it into partial fractions: \[ \frac{1}{u(1+u)} = \frac{A}{u} + \frac{B}{1+u} \] Multiplying by the denominator \( u(1+u) \) gives: \[ 1 = A(1+u) + Bu \] Setting \( u = 0 \) gives \( A = 1 \). Setting \( u = -1 \) gives \( B = -1 \). Now we can rewrite the integral as: \[ -\int \left(\frac{1}{u} - \frac{1}{1+u}\right) du \] Integrating each term: \[ -\left( \ln|u| - \ln|1+u| \right) + C = -\ln\left|\frac{u}{1+u}\right| + C \] Substituting back \( u = \cos(x) \): \[ -\ln\left|\frac{\cos(x)}{1+\cos(x)}\right| + C \] And there you have the final result for the indefinite integral: \[ \int \frac{\sin(x)}{\cos(x) + \cos^2(x)} dx = -\ln\left|\frac{\cos(x)}{1+\cos(x)}\right| + C \]