If I_n = int_{-pi}^{pi} frac{sin(nx)}{(1+pi^x | vedclass

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(A) Given $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx \quad \ldots (i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,

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$(A, B, C)$

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(A) Given $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx \quad \ldots (i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we get$I_n = \int_{-\pi}^{\pi} \frac{\pi^x \sin(nx)}{(1+\pi^x) \sin x} dx \quad \ldots (ii)$Adding $(i)$ and $(ii)$:$2I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{\sin x} dx = 2 \int_0^{\pi} \frac{\sin(nx)}{\sin x} dx$ (since the integrand is an even function).Thus,$I_n = \int_0^{\pi} \frac{\sin(nx)}{\sin x} dx$.Now,$I_{n+2} - I_n = \int_0^{\pi} \frac{\sin((n+2)x) - \sin(nx)}{\sin x} dx = \int_0^{\pi} \frac{2 \cos((n+1)x) \sin x}{\sin x} dx = 2 \int_0^{\pi} \cos((n+1)x) dx = 0$.So,$I_{n+2} = I_n$.For $n=1$,$I_1 = \int_0^{\pi} \frac{\sin x}{\sin x} dx = \pi$.For $n=2$,$I_2 = \int_0^{\pi} \frac{\sin(2x)}{\sin x} dx = \int_0^{\pi} 2 \cos x dx = 0$.Since $I_{n+2} = I_n$,all odd terms $I_{2m+1} = \pi$ and all even terms $I_{2m} = 0$.Therefore,$\sum_{m=1}^{10} I_{2m+1} = 10\pi$ and $\sum_{m=1}^{10} I_{2m} = 0$.Thus,options $(A)$,$(B)$,and $(C)$ are correct.

If $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,then
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$

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If $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,then$(A)$ $I_n = I_{n+2}$$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$$(D)$ $I_n = I_{n+1}$