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.) \( \int \frac{\sin (x)}{\cos (x)+\cos ^{2}(x)} d x \)

We begin with the integral \[ \int \frac{\sin(x)}{\cos(x)+\cos^2(x)}\,dx. \] **Step 1. Factor the denominator.** Notice that we can factor a \(\cos(x)\) from the denominator: \[ \cos(x)+\cos^2(x)= \cos(x)(1+\cos(x)). \] Then the integral becomes \[ \int \frac{\sin(x)}{\cos(x)(1+\cos(x))}\,dx. \] **Step 2. Use a substitution.** Let \[ u=\cos(x). \] Then, differentiating both sides gives \[ \frac{du}{dx}=-\sin(x) \quad \Longrightarrow \quad du = -\sin(x)\,dx. \] Thus \[ \sin(x)\,dx = -du. \] Substitute into the integral: \[ \int \frac{\sin(x)}{\cos(x)(1+\cos(x))}\,dx = -\int \frac{du}{u(1+u)}. \] **Step 3. Decompose into partial fractions.** We express \[ \frac{1}{u(1+u)} \] as a sum of partial fractions: \[ \frac{1}{u(1+u)}=\frac{A}{u}+\frac{B}{1+u}. \] Multiply both sides by \(u(1+u)\): \[ 1 = A(1+u) + Bu. \] This simplifies to \[ 1 = A + (A+B)u. \] Equate the coefficients of like terms: - Constant term: \(A=1\). - Coefficient of \(u\): \(A+B=0\) which implies \(B=-1\). Thus, the decomposition is \[ \frac{1}{u(1+u)}=\frac{1}{u}-\frac{1}{1+u}. \] **Step 4. Integrate each term.** Substitute the partial fractions into the integral: \[ -\int \left(\frac{1}{u}-\frac{1}{1+u}\right)du = -\left(\int \frac{1}{u}\,du - \int \frac{1}{1+u}\,du\right). \] Integrate each part: \[ \int \frac{1}{u}\,du = \ln|u| \quad \text{and} \quad \int \frac{1}{1+u}\,du = \ln|1+u|. \] Thus, \[ -\left(\ln|u| - \ln|1+u|\right) + C = -\ln|u| + \ln|1+u| + C. \] Combine the logarithms: \[ -\ln|u| + \ln|1+u| = \ln\left|\frac{1+u}{u}\right|. \] **Step 5. Substitute back.** Recall that \(u=\cos(x)\). Therefore, the integral becomes \[ \ln\left|\frac{1+\cos(x)}{\cos(x)}\right| + C. \] Thus the final answer is \[ \boxed{\ln\left|\frac{1+\cos(x)}{\cos(x)}\right|+C}. \]