(A) We have the integral

$${I_n} = \int\limits_{ - \pi }^\pi {{{\sin nx} \over {(1 + {\pi ^x})\sin x}}dx} $$

$$ = \int\limits_0^\pi {\left( {{{\sin nx} \over {(1 + {\pi ^x})\sin x}} + {{{\pi ^x}\sin nx} \over {(1 + {\pi ^x})\sin x}}} \right)dx = \int\limits_0^\pi {{{\sin nx} \over {\sin x}}} } $$

Now, $${I_{n + 2}} - {I_n} = \int\limits_0^\pi {{{\sin (n + 2)x - \sin nx} \over {\sin x}}dx} $$

(B) Since $${I_3} = {I_5} = \,...\, = {I_{21}}$$, we have

$$\sum\limits_{m = 1}^{10} {{I_{2m + 1}} = 10{I_3} = 10\int\limits_0^\pi {{{\sin 3x} \over {\sin x}}dx = 10\int\limits_0^\pi {(3 - 4{{\sin }^2}x)dx} } } $$

$$ = 10[3x - 2x + \sin 2x]_0^\pi = 2\pi $$

(C) Since $${I_2} = {I_4} = \,...\, = {I_{20}}$$, we have

$$\sum\limits_{m = 1}^{10} {{I_{2m}} = 10\int\limits_0^\pi {{{\sin 2x} \over {\sin x}}dx = 20[\sin x]_0^\pi = 0} } $$