If m, l, r, s, n are integers such that 9 m | vedclass

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(C) Given $9 > m > l > s > n > r > 2$ ... $(i)$ For $\int_{-2 \pi}^{2 \pi} \sin ^m x \cos ^n x \, dx = 4 \int_0^\pi \sin ^m x \cos ^n x \, dx$, the in

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$(s-2)(s+2) = ln - 3$

Vedclass · Answer

(C) Given $9 > m > l > s > n > r > 2$ ... $(i)$ For $\int_{-2 \pi}^{2 \pi} \sin ^m x \cos ^n x \, dx = 4 \int_0^\pi \sin ^m x \cos ^n x \, dx$, the integral over $[-2\pi, 2\pi]$ is $4 \int_0^{\pi} f(x) dx$ only if $m$ is even. Thus, $m = 8$. For $\int_{-\pi}^\pi \sin ^r x \cos ^s x \, dx = 4 \int_0^{\pi / 2} \sin ^r x \cos ^s x \, dx$, this holds if $r$ is even and $s$ is even. Given the constraints $9 > m > l > s > n > r > 2$ and $m=8$, we have $8 > l > s > n > r > 2$. For $\int_{-\pi / 2}^{\pi / 2} \sin ^l x \cos ^m x \, dx = 0$, the integrand must be an odd function, which implies $l$ is odd. With $m=8$, the remaining integers are $7, 6, 5, 4, 3$. Since $l$ is odd and $l Since $s$ is even and $s Since $n  r > 2$, we have $n=5, r=4$. Checking the options: $(s-2)(s+2) = (6-2)(6+2) = 4 	imes 8 = 32$. $ln - 3 = (7 	imes 5) - 3 = 35 - 3 = 32$. Thus, $(s-2)(s+2) = ln - 3$ is correct.

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