Question

For $n\ge 0$, we have $I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{\sin nx}{(1+2^x)\sin x}dx$
, then

• A. If $n$ is even, then $I_n=0$
• B. If $n$ is odd, then $I_n=\frac{\pi }{2}$
• C. If $n$ is even, then $I_n=\pi$
• D. If $n$ is odd, then $I_n=0$

Answer 1

The correct option is A If $n$ is even, then $I_n=0$

$I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{\sin nx}{(1+2^x)\sin x}dx$ -------(1)
Using $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{\sin nx}{(1+2^{-x})\sin x}dx$ --------(2)
Adding (1) & (2)
$2I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx$
Since $\underset{0}{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx=\underset{-\pi }{\overset{0}{\int }}\frac{\sin nx}{\sin x}dx$
$I_n=\underset{0}{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx$
$I_{n+2}=\underset{0}{\overset{\pi }{\int }}\frac{\sin (n+2)x}{\sin x}dx$
$I_{n+2}-I_n=\underset{0}{\overset{\pi }{\int }}\frac{\sin (n+2)x-\sin nx}{\sin x}dx$
$=2\underset{0}{\overset{\pi }{\int }}\cos (n+1)xdx=0$ for $n\in N$
$I_{n+2}=I_n$
$\begin{array}{c}\text{if}n\text{is even},I_n=I_0=0\\ \text{if}n\text{is odd},I_n=I_1=\pi \end{array}\}$ $\text{As}{I}_n=\underset{0}{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx$