Solve this $$\left[ {\sum\limits_{n = 1}^{10} {\int\limits_{ - 2n - 1}^{ - 2n} {{{\sin }^{27}}x\,dx} } } \right] + \left[ {\sum\limits_{n = 1}^{10} {\int\limits_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} } } \right] = ?$$

Solution :
General term of the series $${\sum\limits_{n = 1}^{10} {\int\limits_{ - 2n - 1}^{ - 2n} {{{\sin }^{27}}x\,dx} } }$$       is
$$\eqalign{ & {I_1} = \int\limits_{ - 2n - 1}^{ - 2n} {{{\sin }^{27}}x\,dx} \cr & = \int\limits_{2n + 1}^{2n} {{{\sin }^{27}}\left( { - x} \right)\left( { - \,dx} \right)} \cr & = - \int\limits_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} \cr & = - {I_2} \cr} $$
Where $${I_2}$$ is general term of series
$$\eqalign{ & \sum\limits_{n = 1}^{10} {\int\limits_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} } \cr & {\text{So, }}{I_1} + {I_2} = 0{\text{ for all }}n \cr} $$