If $I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x ; n=0,1,2$, $\ldots$, then

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Aicharya · Accepted Answer

(a) $ \begin{aligned} & I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^xight) \sin x} d x \ & I_n=\int_{-\pi}^\pi \frac{\pi^x \sin n x}{\left(1+\pi^xight) \sin x} d x \ & {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d xight]} \ & 2 I_n=\int_{-\pi}^\pi \frac{\sin n x}{\sin x} d x \end{aligned} $ $ \begin{aligned} & \Rightarrow \quad 2 I_n=2 \int_0^\pi \frac{\sin n x}{\sin x} d x \ & \Rightarrow \quad I_n=\int_0^\pi \frac{\sin n x}{\sin x} d x \ & 	ext { Now, } \quad I_{n+2}-I_n \ & =\int_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x} d x \ & =\int_0^\pi \frac{2 \cos (n+1) x \sin x}{\sin x} d x \ & =2\left[\frac{\sin (n+1) x}{(n+1)}ight]_0^\pi=0 \ & \Rightarrow \quad I_{n+2}=I_n \ & 	ext { (b) As } I_3=I_5=\ldots=I_{21} \ & 	herefore \sum_{m=1}^{10} I_{2 m+1}=10 I_3=10 \int_0^\pi \frac{\sin 3 x}{\sin x} d x \ & =10 \int_0^\pi\left(3-4 \sin ^2 xight) d x \ & =10[3 x-2 x+\sin 2 x]_0^\pi=10 \pi \ & 	ext { (c) As } \quad I_2=I_4=\ldots=I_{20} \ & 	herefore \quad \sum_{m=1}^{10} I_{2 m}=10 \int_0^\pi \frac{\sin 2 x}{\sin x} d x \ & =20[\sin x]_0^\pi=0 \ & \end{aligned} $

If $I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x ; n=0,1,2$, $\ldots$, then

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