Official solution
\[\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta \]
\[ = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta \]
Put \[1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t\]
So \[{{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt\]
Now \[{1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt} \]
\[ = {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C\]
\[ = {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C\]