$\cfrac{1}{{\sqrt {{{\sin }^3}x\sin \left( {x + \alpha } \right)} }}$

: Let $I = \int {\cfrac{1}{{\sqrt {{{\sin }^3}x\sin \left( {x + \alpha } \right)} }}dx} = \int {\sqrt {\cfrac{{\sin x}}{{\sqrt {{{\sin }^4}x\sin \left( {x + \alpha } \right)} }}} } dx$
$= \int {\cfrac{1}{{{{\sin }^2}x}}\sqrt {\cfrac{{\sin x}}{{\sin \left( {x + \alpha } \right)}}} } dx$

Let $\cfrac{{\sin \left( {x + \alpha } \right)}}{{\sin x}} = t$

$\Rightarrow$ $\cfrac{{\sin x\cos \left( {x + \alpha } \right) - \cos \,x\sin \left( {x + \alpha } \right)}}{{{{\sin }^2}x}}dx = dt$

$\Rightarrow$ $\cfrac{{\sin \left[ {x - \left( {x + \alpha } \right)} \right]}}{{{{\sin }^2}x}}dx = dt$ $\Rightarrow$ $- \cfrac{{\sin \alpha }}{{{{\sin }^2}x}}dx = dt$

$\therefore$ $I = \int { - \cfrac{1}{{\sin \alpha }} \cdot \cfrac{1}{{\sqrt t }}dt = - \cfrac{1}{{\sin \alpha }}} \int {{t^{ - 1/2}}dt}$

$= - \cfrac{1}{{\sin \alpha }}\left( {\cfrac{{{t^{1/2}}}}{{1/2}}} \right) + C = \cfrac{{ - 2}}{{\sin \alpha }}\sqrt t + C = \cfrac{{ - 2}}{{\sin \alpha }}\sqrt {\cfrac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} + C$