Consider the function $$f(x) = \frac{\sin x}{3 - 2\cos x} \quad (0 \leqq x \leqq \pi)$$ (1) The derivative of $f(x)$ is $$f'(x) = \frac{\mathbf{A}\cos x - \mathbf{B}}{(\mathbf{C} - \mathbf{D}\cos x)^2}.$$ Let $\alpha$ be the value of $x$ at which $f(x)$ has a local extremum. Then we have $$\cos\alpha = \frac{\mathbf{E}}{\mathbf{F}}.$$ (2) The portion of the plane bounded by the graph of the function $y = f(x)$ and the $x$-axis is divided into two parts by the straight line $x = \alpha$. Let $S_1$ be the area of the part located on the left side of the line. Then we have $$S_1 = \int_{\frac{\mathbf{G}}{\mathbf{H}}}^{\mathbf{I}} \frac{dt}{\mathbf{J} - \mathbf{K}t} = \frac{\mathbf{L}}{\mathbf{L}}\log\frac{\mathbf{L}}{\mathbf{L}}.$$ Let $S_2$ be the area of the part located on the right side. We have $$S_2 = \frac{\mathbf{P}}{2}\log\mathbf{Q}.$$
Consider the function
$$f(x) = \frac{\sin x}{3 - 2\cos x} \quad (0 \leqq x \leqq \pi)$$
(1) The derivative of $f(x)$ is
$$f'(x) = \frac{\mathbf{A}\cos x - \mathbf{B}}{(\mathbf{C} - \mathbf{D}\cos x)^2}.$$
Let $\alpha$ be the value of $x$ at which $f(x)$ has a local extremum. Then we have
$$\cos\alpha = \frac{\mathbf{E}}{\mathbf{F}}.$$
(2) The portion of the plane bounded by the graph of the function $y = f(x)$ and the $x$-axis is divided into two parts by the straight line $x = \alpha$. Let $S_1$ be the area of the part located on the left side of the line. Then we have
$$S_1 = \int_{\frac{\mathbf{G}}{\mathbf{H}}}^{\mathbf{I}} \frac{dt}{\mathbf{J} - \mathbf{K}t} = \frac{\mathbf{L}}{\mathbf{L}}\log\frac{\mathbf{L}}{\mathbf{L}}.$$
Let $S_2$ be the area of the part located on the right side. We have
$$S_2 = \frac{\mathbf{P}}{2}\log\mathbf{Q}.$$