I_n = ∫_(−π) ^( π) (( sin(nx ))/((1 + e^x )sinx)) dx =?


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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
	$I_n =∫_(−π) ^π ((sin nx)/((e^x +1)sin x))dx =^([t=−x]) =−∫_π ^(−π) ((e^t sin nt)/((e^t +1)sin t))dt =^([t=x]) ∫_(−π) ^π ((e^x sin nx)/((e^x +1)sin x))dx ⇒ 2I_n =∫_(−π) ^π ((sin nx)/((e^x +1)sin x))dx+∫_(−π) ^π ((e^x sin nx)/((e^x +1)sin x))dx= =∫_(−π) ^π ((sin nx)/(sin x))dx ⇒ I_n = { ((0; n=2k+1)),((π; n=2k)) :}$
	$I_{n} = \underset{- π}{\int^{π}} \frac{\sin n x}{(e^{x} + 1) \sin x} d x \overset{[t = - x]}{=}$ $= - \underset{π}{\int^{- π}} \frac{e^{t} \sin n t}{(e^{t} + 1) \sin t} d t \overset{[t = x]}{=} \underset{- π}{\int^{π}} \frac{e^{x} \sin n x}{(e^{x} + 1) \sin x} d x$ $\Rightarrow$ $2 I_{n} = \underset{- π}{\int^{π}} \frac{\sin n x}{(e^{x} + 1) \sin x} d x + \underset{- π}{\int^{π}} \frac{e^{x} \sin n x}{(e^{x} + 1) \sin x} d x =$ $= \underset{- π}{\int^{π}} \frac{\sin n x}{\sin x} d x$ $\Rightarrow$ $I_{n} = {\begin{cases} 0; n = 2 k + 1 \\ π; n = 2 k \end{cases}$
	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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