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

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Professor PC Thomas Classes · Answer

Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$, we find $I_n = \int_{0}^{\pi} \frac{\sin nx}{\sin x} dx$. Now, $I_{n+2} - I_n = \int_{0}^{\pi} \frac{\sin(n+2)x - \sin nx}{\sin x} dx = \int_{0}^{\pi} \frac{2\cos(n+1)x \sin x}{\sin x} dx = 2\int_{0}^{\pi} \cos(n+1)x dx = 0$. Thus, $I_{n+2} = I_n$ (Option A is correct). We know $I_0 = 0$, so $I_0 = I_2 = I_4 = \dots = 0$. Hence $\sum_{m=1}^{10} I_{2m} = 0$ (Option C is correct). Also, $I_1 = \int_{0}^{\pi} 1 dx = \pi$, so $I_1 = I_3 = I_5 = \dots = \pi$. Therefore, $\sum_{m=1}^{10} I_{2m+1} = 10 	imes \pi = 10\pi$ (Option B is correct).