The value of sum_{n=1}^{10} int_{-2n-1}^{-2n} | vedclass

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(D) Let $I = \sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$. Consider the first integral t

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(D) Let $I = \sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$. Consider the first integral term: $J_n = \int_{-2n-1}^{-2n} \sin^{27} x \, dx$. Let $x = -t$, then $dx = -dt$. When $x = -2n-1$, $t = 2n+1$. When $x = -2n$, $t = 2n$. So, $J_n = \int_{2n+1}^{2n} \sin^{27}(-t) (-dt) = \int_{2n}^{2n+1} -\sin^{27} t \, dt = -\int_{2n}^{2n+1} \sin^{27} t \, dt$. Substituting this back into the sum, we get: $I = \sum_{n=1}^{10} (-\int_{2n}^{2n+1} \sin^{27} x \, dx) + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$. $I = -\sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx = 0$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

The value of $\sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$ is equal to

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