SATHEE: Definite Integration Question 52

Definite Integration Question 52

Question 52

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) $\sum_{m = 1}^{10} I_{2 m + 1} = 10 π$ (c) $\sum_{m = 1}^{10} I_{2 m} = 0$ (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

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

$$ \begin{aligned} & =2 \int_{0}^{\pi} \cos (n+1) x d x=2 \frac{\sin (n+1) x}{(n+1)}{ }{0}^{\pi}=0 \ & \therefore \quad I{n+2}=I_{n} \end{aligned} $$

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

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

Definite Integration Question 52

Question 52

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Definite Integration Question 52

Question 52

Numerical Value

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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