Let $g(x) = \int \frac{1 + 2\cos x}{(\cos x + 2)^2} dx$ and $g(0) = 0$, then the value of $8g(\pi/2)$ is
Step-by-Step Solution
Key Concept: Splitting a complex fraction and recognizing that it can be expressed as the derivative of $\frac{\sin x}{\cos x + 2}$ through careful algebraic work.
To find $g(x) = \int \frac{\cos x(\cos x + 2) + \sin^2 x}{(\cos x + 2)^2}dx$, split the numerator to get $\int \frac{\cos x}{(\cos x+2)}dx + \int \frac{\sin^2 x}{(\cos x+2)^2}dx$. Using substitution and algebraic manipulation, this simplifies to $g(x) = \frac{\sin x}{\cos x + 2} + C$. Applying the initial condition $g(0) = 0$ gives $C = 0$.
Correct Answer: 1