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

Let u = cosθ, du = -sinθ dθ. Integral becomes -∫ du/[(2+u)(3+4u)]. Partial fractions: 1/[(2+u)(3+4u)] = A'/(2+u) + B'/(3+4u). Solving: 1 = A'(3+4u) + B'(2+u). At u = -2: 1 = A'(-5), A' = -1/5. At u = -3/4: 1 = B'(5/4), B' = 4/5. So integral = -[(-1/5)ln|2+u| + (4/5)(1/4)ln|3+4u|] = (1/5)ln|2+cosθ| - (1/5)ln|3+4cosθ|. Hence A = 1/5.