Back to our solution.
\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\int_{\beta}^{a}f(s)\cos ks \ ds + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
Substitute the last result we obtained.
\(\displaystyle y(x) = \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\left[\frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) \right] + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
The expression (solution) looks very complicated and I still don't know if my method works.

Let us try to simplify and see what happens.
Let us start by simplifying the brackets: \(\displaystyle \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right)\)
\(\displaystyle \left(\frac{\sin kx}{k} - \frac{\cos kx\sin ka}{k\cos ka}\right) = \left(\frac{\sin kx\cos ka}{k\cos ka} - \frac{\cos kx\sin ka}{k\cos ka}\right) = \frac{\sin k(x - a)}{k\cos ka}\)
This gives:
\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\cos ka}\left[\frac{\cos ka}{\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) \right] + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{b}f(s)\sin ks \ ds\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
\(\displaystyle y(x) = \frac{\sin k(x - a)}{k\sin k(a - b)}\left(\sin kb\int_{a}^{x}f(s)\cos ks \ ds + \sin kb\int_{x}^{b}f(s)\cos ks \ ds - \cos kb\int_{a}^{x}f(s)\sin ks \ ds - \cos kb\int_{x}^{b}f(s)\sin ks\right) + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks + \frac{\sin kx}{k}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx}{k}\int_{a}^{x}f(s)\sin ks \ ds\)
\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks \ ds + \frac{\sin kx\sin ka\cos kb - \sin kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka\cos kb - \cos kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds\)
\(\displaystyle y(x) = \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin k(x - a)\sin kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\cos ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds - \frac{\sin k(x - a)\cos kb}{k\sin k(a - b)}\int_{x}^{b}f(s)\sin ks \ ds - \frac{\sin kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds + \frac{\sin kx\sin ka\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\cos ks \ ds - \frac{\cos kx\sin ka\cos kb}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds + \frac{\cos kx\sin kb\cos ka}{k\sin k(a - b)}\int_{a}^{x}f(s)\sin ks \ ds\)
Ahhhhhh that's tough!

@khansaheb
Can you check where I went wrong?