left[ {sumlimits_{n = 1}^{10} {int_{ - 2n - 1 | vedclass

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(D) Let the given expression be $S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \sum\limits_{n = 1}^{10} \int_{2n}^{2n + 1} \si

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(D) Let the given expression be $S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \sum\limits_{n = 1}^{10} \int_{2n}^{2n + 1} \sin^{27}x \, dx$. Consider the first summation: $I_1 = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx$. Let $x = -t$,then $dx = -dt$. When $x = -2n-1$,$t = 2n+1$. When $x = 2n$,$t = -2n$. Thus,$\int_{ - 2n - 1}^{2n} \sin^{27}x \, dx = \int_{2n+1}^{-2n} \sin^{27}(-t) (-dt) = \int_{2n+1}^{-2n} -\sin^{27}t (-dt) = \int_{2n+1}^{-2n} \sin^{27}t \, dt = -\int_{-2n}^{2n+1} \sin^{27}t \, dt$. This approach is complex. Let's simplify the sum: $S = \sum\limits_{n = 1}^{10} \left( \int_{ - 2n - 1}^{2n} \sin^{27}x \, dx + \int_{2n}^{2n + 1} \sin^{27}x \, dx \right)$. By the property of definite integrals,$\int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx = \int_{a}^{c} f(x) dx$. So,$S = \sum\limits_{n = 1}^{10} \int_{ - 2n - 1}^{2n + 1} \sin^{27}x \, dx$. Since $\sin^{27}x$ is an odd function,$\int_{-a}^{a} \sin^{27}x \, dx = 0$. Therefore,$\int_{ - (2n + 1)}^{(2n + 1)} \sin^{27}x \, dx = 0$ for each $n$. Thus,$S = \sum\limits_{n = 1}^{10} 0 = 0$.

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

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