If $a=2 n$ and $b=2 m+1$ for all $m, n \in \mathbb{N}$, $$ \int_{-\pi}^\pi e^{\sin ^a x} \cot ^b(2 n+1) x d…
If $a=2 n$ and $b=2 m+1$ for all $m, n \in \mathbb{N}$,
$$
\int_{-\pi}^\pi e^{\sin ^a x} \cot ^b(2 n+1) x d x=
$$
$0$
$1$
$-1$
$\pi$
Solution
Given $a=2 n, b=2 m+1$ and $m, n \in \mathrm{N}$
Let $I=\int_{-\pi}^\pi e^{\sin ^a x} \cot ^b(2 n+1) x d x$
Let $f(x)=e^{\sin ^a x} \cdot \cot ^b(2 n+1) x$
$
\Rightarrow \quad f(-x)=e^{\sin ^a(-x)} \cot ^b(2 n+1)(-x)
$
Since $a$ is even number and $b$ is odd number and $\cot (-\theta)=-\cot \theta$
$
\begin{aligned}
& \Rightarrow f(-x)=e^{\sin ^a(x)} \cdot\left[-\cot ^b(2 n+1) x\right] d x \\
& \Rightarrow f(-x)=-f(x)
\end{aligned}
$
Hence
$
I=\int_{-\pi}^\pi f(x) d x=0\left\{\begin{array}{l}
\because f(-x)=-f(x) \\
\int_{-a}^a f(x) d x=0
\end{array}\right.
$