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Question: \(\sum _ { n = 1 } ^ { 10 } \int _ { - 2 n - 1 } ^ { - 2 } \sin ^ { 27 } x d x\) +<img src="https://...

∑n=110∫−2n−1−2sin⁡27xdx\sum _ { n = 1 } ^ { 10 } \int _ { - 2 n - 1 } ^ { - 2 } \sin ^ { 27 } x d x +equals -

A

27

B

54

C

–54

D

0

Answer

0

Explanation

Solution

let I=∫−2n−12nsin⁡27πI = \int _ { - 2 n - 1 } ^ { 2 n } \sin ^ { 27 } \pi

I1 = ∫2n+12nsin⁡27(−t)(−dt)−dx=dt−x=t\int _ { 2 \mathrm { n } + 1 } ^ { 2 \mathrm { n } } \sin ^ { 27 } ( - \mathrm { t } ) ( - \mathrm { dt } ) _ { - \mathrm { dx } = \mathrm { dt } } ^ { - \mathrm { x } = \mathrm { t } }

= −∫2n2n+1sin⁡27+dt- \int _ { 2 n } ^ { 2 n + 1 } \sin ^ { 27 } + \mathrm { dt } =

̃ −∫−2n−1−2nsin⁡27x+∫2n2n+1sin⁡27xdx- \int _ { - 2 n - 1 } ^ { - 2 n } \sin ^ { 27 } x + \int _ { 2 n } ^ { 2 n + 1 } \sin ^ { 27 } x d x = 0