Exponential and Trigonometry in Integrand

$I= \displaystyle \int_{-\pi}^{\pi}\dfrac{\sin(nx)}{(1+2^{x})\sin x} dx$

$\displaystyle \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b - x) dx$

$\displaystyle\int_{-\pi}^{\pi}\dfrac{2^x \sin (nx)}{(1+2^{x})\sin x} dx$

$2I = \displaystyle \int_{-\pi}^{\pi} \dfrac{\sin(nx)[1 + 2^x]}{(1+2^{x})\sin x} dx$

$2I = 2\displaystyle \int_{0}^{\pi} \dfrac{\sin (nx)}{\sin x}$

$n~is~an~odd~nummber \implies n = 2k + 1 , k \in I^{+}$

$I = \displaystyle \int_{0}^{\pi} \dfrac{\sin (2k + 1)x}{\sin x}$

$I = \displaystyle \int_{0}^{\pi} \dfrac{ \sin 2kx \times \cos x}{\sin x} + \dfrac{ \cos nx \times \sin x}{ \sin x} dx$

$I = \displaystyle \int_{0}^{\pi} \dfrac{\sin 2k x\times \cos x}{\sin x} + \cos nx dx$

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$\dfrac{ \sin 2k }{\sin x} = 2( \cos x + \cos 3x + ..... + \cos (2k - 1)x)$

Proof-

$\sin 2kx = 2\sin x(\cos x + \cos 3x + ..... + \cos (2k - 1)x)$

$\sin 2kx = \sin 2x + \sin 4x - \sin 2x + \sin 6x - \sin 4x + .... + \sin 2kx - \sin(2k - 2)x)$

$\sin 2kx = \sin 2kx$

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$I = \displaystyle \int_{0}^{\pi} 2\cos x( \cos x + \cos 3x + ..... + \cos (2k - 1)x) + \cos nx dx$

$I = \displaystyle \int_{0}^{\pi} 1 + \cos 2x + \cos 4x + \cos 2x + .... + \cos 2kx + \cos (2k - 2)x + (\cos nx) dx$

$I = \pi$ $\left ( \left[ \dfrac{\sin nx}{n} \right]_{0}^{\pi} = 0\right )$

Just set n=1 and its trivial