If a=2n and b=2m+1 for all m, n in N,then eva | vedclass

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(A) Given $a=2n$ and $b=2m+1$ where $m, n \in N$. Let $I = \int_{-\pi}^{\pi} f(x) dx$,where $f(x) = e^{\sin^a x} \cdot \cot^b((2n+1)x)$. We check the

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(A) Given $a=2n$ and $b=2m+1$ where $m, n \in N$. Let $I = \int_{-\pi}^{\pi} f(x) dx$,where $f(x) = e^{\sin^a x} \cdot \cot^b((2n+1)x)$. We check the parity of the function $f(x)$: $f(-x) = e^{\sin^a(-x)} \cdot \cot^b((2n+1)(-x))$. Since $a=2n$ is an even number,$\sin^a(-x) = (\sin(-x))^a = (-\sin x)^a = \sin^a x$. Since $b=2m+1$ is an odd number,$\cot^b((2n+1)(-x)) = (\cot(-(2n+1)x))^b = (-\cot((2n+1)x))^b = -\cot^b((2n+1)x)$. Therefore,$f(-x) = e^{\sin^a x} \cdot (-\cot^b((2n+1)x)) = -f(x)$. Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi, \pi]$ is zero. Thus,$I = 0$.

If $a=2n$ and $b=2m+1$ for all $m, n \in N$,then evaluate the integral: $\int_{-\pi}^{\pi} e^{\sin^a x} \cot^b((2n+1)x) dx$.

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