Let ∫ sinθ dθ / [(2+cosθ)(3+4cosθ)] = A ln|2+cosθ| + B ln|3+4cosθ|. What is the value of A?

Let u = cosθ, then du = -sinθ dθ. The integral becomes -∫ du/[(2+u)(3+4u)]. Using partial fractions: 1/[(2+u)(3+4u)] = A'/(2+u) + B'/(3+4u). Solving: 1 = A'(3+4u) + B'(2+u). Setting u = -2: 1 = A'(-5), so A' = -1/5. Setting u = -3/4: 1 = B'(5/4), so B' = 4/5. After integration with the negative sign: A = 1/5·... Working it through carefully, A = -1/5.