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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 ImI_m equal to?

Options

  1. A0
  2. B1
  3. Cmm
  4. D2m2m

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}=∫0πsin⁡2m(π−x)sin⁡(π−x)dx\begin{aligned} & =\int_0^\pi \frac{\sin 2 m(\pi-x)}{\sin (\pi-x)} d x \end{aligned}=∫0πsin⁡(2mπ−2mx)sin⁡xdx=∫0π−sin⁡2mxsin⁡xdx=\int_0^\pi \frac{\sin (2 m \pi-2 m x)}{\sin x} d x=\int_0^\pi \frac{-\sin 2 \mathrm{mx}}{\sin \mathrm{x}} \mathrm{dx}Im=−∫0ππsin⁡2mxsin⁡xdx2Im=0⇒Im=0\begin{aligned} & \mathrm{I}_{\mathrm{m}}=-\int_0^\pi \frac{\pi \sin 2 \mathrm{mx}}{\sin \mathrm{x}} \mathrm{dx} \\ & 2 \mathrm{I}_{\mathrm{m}}=0 \Rightarrow \mathrm{I}_{\mathrm{m}}=0\end{aligned}$$
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