Official solution
\[\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} \]
= \[\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx \]
= \[\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx\]
= \[\int {{{\sin 3x + \sin 2x} \over {\sin x}}} dx\]
Use sin2x = 2sinxcosx and sin3x = 3sinx
-- 4sin3x
= \[\int {{{3\sin x - 4{{\sin }^3}x + 2\sin x\cos x} \over {\sin x}}} dx\]
= \[\int {\left( {3 - 4{{\sin }^2}x + 2\cos x} \right)} dx\]
= \[\int {\left( {3 - 2\left( {1 - \cos 2x} \right) + 2\cos x} \right)} dx\]
= \[\int {\left( {1 + 2\cos 2x + 2\cos x} \right)} dx\]
= x + sin2x + 2sinx + C