I-n-pi-pi-sin-nx-1-e-x-sinx-dx-

Question Number 212139 by mnjuly1970 last updated on 03/Oct/24

I_{n} = \int_{- π}^{π} \frac{\sin (n x)}{(1 + e^{x}) \sin x} d x = ?

Answered by Frix last updated on 03/Oct/24

I_{n} = \underset{- π}{\int^{π}} \frac{\sin n x}{(e^{x} + 1) \sin x} d x \overset{[t = - x]}{=}

= - \underset{π}{\int^{- π}} \frac{e^{t} \sin n t}{(e^{t} + 1) \sin t} d t \overset{[t = x]}{=} \underset{- π}{\int^{π}} \frac{e^{x} \sin n x}{(e^{x} + 1) \sin x} d x

\Rightarrow

2 I_{n} = \underset{- π}{\int^{π}} \frac{\sin n x}{(e^{x} + 1) \sin x} d x + \underset{- π}{\int^{π}} \frac{e^{x} \sin n x}{(e^{x} + 1) \sin x} d x =

= \underset{- π}{\int^{π}} \frac{\sin n x}{\sin x} d x

\Rightarrow

I_{n} = {\begin{cases} 0; n = 2 k + 1 \\ π; n = 2 k \end{cases}

Commented by mnjuly1970 last updated on 03/Oct/24

t h a n k s a l o t s i r F r i x

Commented by Frix last updated on 03/Oct/24

{You}^{'} re welcome!

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