I'm having trouble solving part (2). How do I do it? I don't get where the integrations numbers came from (2/3,1) and this new equation for integration too. How do I find the value of $x = \alpha$ if it's not an rational angle? I want to understand how I can comprehend and reach the answer myself.
The original question, part (1) has been solved:
Consider $$f(x)=\frac{\sin x}{(3-2\cos x)}$$ when $0\leq x \leq \pi$
$(1):$ $$f'(x) = \frac{3\cos x-2}{(3-2\cos x)^2}$$
Let $ \alpha$ be the value of $x$ at which $f(x)$ has a local extremum. Then we have $\cos\alpha = \frac{2}{3}$
$(2):$
The portion of the plane bounded by the graph of the function $y = f(x)$ and the $X$ axis is divided into $2$ parts by the straight line $x = \alpha$. Let $S_1$ be the area of the part located at the left side of the line. Then we have $$S_1 = \int_{\frac{G}{H}}^{I} \frac{dt}{J-Kt}=\frac{L}{N}\log\left(\frac{M}{O}\right)$$ Let $S_2$ be the area of the part located on the right side. We have $$S_2=\frac{P}{2}\log(Q)$$
The letters stand for $G=2, H=3, I=1, J=3, K=2, L=1, M=2, N=5, O=3, P=1, Q=3.$
$$S_1 = \int_0^{\cos^{-1}{\frac{2}{3}}} \frac{\sin x}{3-2\cos x}dx$$
Let $\cos x = t$
$\implies -\sin x dx = dt$
When $x=0 \to t = 1$ and $x=\arccos \frac{2}{3} \to t = \frac{2}{3}$ using our subsitution above to define the limits of $t$ after the change in variable
$$S_1 = -\int_1^{\frac{2}{3}}\frac{dt}{3-2t} = \int_{\frac{2}{3}}^1 \frac{dt}{3-2t}$$
Similarly for $S_2$