SATHEE: Indefinite Integration 1 Question 52

Indefinite Integration 1 Question 52

53. If $I_{n} = \int_{- π}^{π} \frac{\sin n x}{(1 + π^{x}) \sin x} d x, n = 0, 1, 2, \dots$ , then

(a) $I_{n} = I_{n + 2}$

(b) $\sum_{m = 1}^{10} I_{2 m + 1} = 10 π$

(d) $I_{n} = I_{n + 1}$

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Solution:

Given

$I_{n} = \int_{- π}^{π} \frac{\sin n x}{(1 + π^{x}) \sin x} d x$

Using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (b + a - x) d x$ , we get

$I_{n} = \int_{- π}^{π} \frac{π^{x} \sin n x}{(1 + π^{x}) \sin x} d x$

On adding Eqs. (i) and (ii), we have

$\begin{aligned} 2 I_{n} = \int_{- π}^{π} \frac{\sin n x}{\sin x} d x = 2 \int_{0}^{π \sin n x} \frac{\sin x}{\sin} d x \\ [∵ f (x) = \frac{\sin n x}{\sin x} is an even function] \\ \Rightarrow I_{n} = \int_{0}^{π} \frac{\sin n x}{\sin x} d x \\ Now, I_{n + 2} - I_{n} = \int_{0}^{π} \frac{\sin (n + 2) x - \sin n x}{\sin x} d x \end{aligned}$

$\begin{aligned} = 2 \int_{0}^{π} \cos (n + 1) x d x = 2 \frac{\sin (n + 1) x}{(n + 1)}_{0}^{π} = 0 \\ ∴ I_{n + 2} = I_{n} \end{aligned}$

$\Rightarrow \sum_{m = 1}^{10} I_{2 m + 1} = 10 π$ and $\sum_{m = 1}^{10} I_{2 m} = 0$

$∴$ Correct options are (a), (b), (c).

Indefinite Integration 1 Question 52

53. If $I_{n} = \int_{- π}^{π} \frac{\sin n x}{(1 + π^{x}) \sin x} d x, n = 0, 1, 2, \dots$ , then

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Solution:

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Indefinite Integration 1 Question 52

53. If In=∫−ππsin⁡nx(1+πx)sin⁡xdx,n=0,1,2,…, then

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53. If $I_{n} = \int_{- π}^{π} \frac{\sin n x}{(1 + π^{x}) \sin x} d x, n = 0, 1, 2, \dots$ , then