\[\left[ {\,\mathop \sum \limits_{n = 1}^{10} \,\,\mathop \smallint \limits_{ - 2n - 1}^{ - 2n} \,\,{{\sin }^{27}}\,\,x\,dx} \right]\,\, + \,\,\left[ {\mathop \sum \limits_{n = 1}^{10} \,\,\mathop \smallint \limits_{2n}^{2n + 1} \,\,{{\sin }^{27}}\,\,x\,dx} \right]\] equals

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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\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]

equals

View answer and explanation

Correct option: D

\mathrm{0}

Option D (0) is correct because the sum of the integrals can be shown to cancel out by using the substitution x = -t and the properties of the sine function.

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[*]

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