Indefinite Integration 1 Question 52
53. 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
(a) $I _n=I _{n+2}$
(b) $\sum _{m=1}^{10} I _{2 m+1}=10 \pi$
(c) $\sum _{m=1}^{10} I _{2 m}=0$
(d) $I _n=I _{n+1}$
Numerical Value
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Solution:
- Given
$$ I _n=\int _{-\pi}^{\pi} \frac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x $$
Using $\int _a^{b} f(x) d x=\int _a^{b} f(b+a-x) d x$, we get
$$ I _n=\int _{-\pi}^{\pi} \frac{\pi^{x} \sin n x}{\left(1+\pi^{x}\right) \sin x} d x $$
On adding Eqs. (i) and (ii), we have
$$ \begin{aligned} & 2 I _n=\int _{-\pi}^{\pi} \frac{\sin n x}{\sin x} d x=2 \int _0^{\pi \sin n x} \frac{\sin x}{\sin } d x \\ & {\left[\because f(x)=\frac{\sin n x}{\sin x} \text { is an even function }\right]} \\ & \Rightarrow \quad I _n=\int _0^{\pi} \frac{\sin n x}{\sin x} d x \\ & \text { Now, } I _{n+2}-I _n=\int _0^{\pi} \frac{\sin (n+2) x-\sin n x}{\sin x} d x \end{aligned} $$
$$ \begin{aligned} & =2 \int _0^{\pi} \cos (n+1) x d x=2 \frac{\sin (n+1) x}{(n+1)}{ } _0^{\pi}=0 \\ & \therefore \quad I _{n+2}=I _n \end{aligned} $$
$\Rightarrow \quad \sum _{m=1}^{10} I _{2 m+1}=10 \pi$ and $\quad \sum _{m=1}^{10} I _{2 m}=0$
$\therefore$ Correct options are (a), (b), (c).
