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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Definite Integration

NDA 2015 · Mathematics · Definite Integration

Consider the integral I_m=∫_0^π ( 2 m x)/( x) d x , where m is a...

Consider the integral Im=∫0πsin⁡2mxsin⁡xdxI_m=\int_0^\pi \frac{\sin 2 m x}{\sin x} d x, where mm is a positive integer. What is I1\mathrm{I}_1 equal to?

Options

  1. A0
  2. B12\frac{1}{2}
  3. C1
  4. D2

Correct answer

A. 0

Step-by-step solution

Im=∫0πsin⁡2mxsin⁡xdx∴\mathrm{I}_{\mathrm{m}}=\int_0^\pi \frac{\sin 2 \mathrm{mx}}{\sin \mathrm{x}} \mathrm{dx}\therefore \quad\mathrm{I}1=∫0πsin⁡2xsin⁡xdx=∫0π2sin⁡xcos⁡xsin⁡xdx_1=\int_0^\pi \frac{\sin 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}=\int_0^\pi \frac{2 \sin \mathrm{x} \cos \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}=2∫0πcos⁡xdx=2[sin⁡x]0π=2[sin⁡π−sin⁡0]=2(0)=0\begin{aligned} & =2 \int_0^\pi \cos \mathrm{xdx}=2[\sin \mathrm{x}]_0^\pi=2[\sin \pi-\sin 0] \\ & =2(0)=0 \end{aligned}∴\therefore \quad Option (A) is correct.
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