IfIn=∫−ππ sin⁡nx1+πxsin⁡xdx,=0,1,2then

In=∫−ππ sin⁡nx1+πxsin⁡xdx=∫0π sin⁡nx1+πxsin⁡x+πxsin⁡nx1+πxsin⁡xdx=∫0π sin⁡nxsin⁡xdx Now, In+2−In=∫0π sin⁡(n+2)x−sin⁡nxsin⁡xdx=∫0π 2cos⁡(n+1)x⋅sin⁡xsin⁡xdx=0⇒I1=π,I2=∫0π 2cos⁡xdx=0

IfIn=∫−ππ sin⁡nx1+πxsin⁡xdx,=0,1,2then

In=∫−ππ sin⁡nx1+πxsin⁡xdx=∫0π sin⁡nx1+πxsin⁡x+πxsin⁡nx1+πxsin⁡xdx=∫0π sin⁡nxsin⁡xdx Now, In+2−In=∫0π sin⁡(n+2)x−sin⁡nxsin⁡xdx=∫0π 2cos⁡(n+1)x⋅sin⁡xsin⁡xdx=0⇒I1=π,I2=∫0π 2cos⁡xdx=0

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If $I_{n} = \int_{- π}^{π} \frac{\sin nx}{(1 + π^{x}) \sin x} dx, = 0, 1, 2$ then

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$I_{n} = I_{n + 2}$

$\sum_{m = 1}^{10} I_{2 m + 1} = 10 π$

$\sum_{m = 1}^{10} I_{2 m} = 0$

$I_{n} = I_{n + 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} I_{n} = \int_{- π}^{π} \frac{\sin nx}{(1 + π^{x}) \sin x} dx \\ = \int_{0}^{π} (\frac{\sin nx}{(1 + π^{x}) \sin x} + \frac{π^{x} \sin nx}{(1 + π^{x}) \sin x}) dx = \int_{0}^{π} \frac{\sin nx}{\sin x} dx \\ Now, I_{n + 2} - I_{n} = \int_{0}^{π} \frac{\sin (n + 2) x - \sin nx}{\sin x} dx \end{array}$
$\begin{array}{l} = \int_{0}^{π} \frac{2 \cos (n + 1) x \cdot \sin x}{\sin x} dx = 0 \\ \Rightarrow I_{1} = π, I_{2} = \int_{0}^{π} 2 \cos xdx = 0 \end{array}$