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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 I2+I3I_2+I_3 equal to?

Options

  1. A4
  2. B2
  3. C1
  4. D0

Correct answer

D. 0

Step-by-step solution

I2=∫0πsin⁡4xsin⁡xdx=∫0π2sin⁡2xcos⁡2xsin⁡xdx=2∫0π2sin⁡xcos⁡xcos⁡2xsin⁡xdx=4∫0πcos⁡x(1−2sin⁡2x)dx=4∫0πcos⁡xdx−8∫0πcos⁡xsin⁡2xdx=0\begin{aligned} & I_2=\int_0^\pi \frac{\sin 4 x}{\sin x} d x=\int_0^\pi \frac{2 \sin 2 x \cos 2 x}{\sin x} d x \\ & =2 \int_0^\pi \frac{2 \sin \mathrm{x} \cos \mathrm{x} \cos 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx} \\ & =4 \int_0^\pi \cos x\left(1-2 \sin ^2 x\right) d x \\ & =4 \int_0^\pi \cos x d x-8 \int_0^\pi \cos x \sin ^2 x d x=0 \end{aligned} Let sin⁡x=tcos⁡xdx=dt=4(sin⁡π−sin⁡0)−0=0I3=∫0πsin⁡6xsin⁡xdx=∫0π2[3sin⁡x−4sin⁡3x]cos⁡3xsin⁡xdx\begin{aligned} & \text{ Let } \sin \mathrm{x}=\mathrm{t} \\ & \cos \mathrm{xdx}=\mathrm{dt} \\ & =4(\sin \pi-\sin 0)-0=0 \\ & I_3=\int_0^\pi \frac{\sin 6 x}{\sin \mathrm{x}} d x \\ & =\int_0^\pi \frac{2\left[3 \sin x-4 \sin ^3 x\right] \cos 3 x}{\sin x} d x \end{aligned}=∫0π2(3−4sin⁡2x)cos⁡3xdx=∫0π6cos⁡3xdx−∫0π8sin⁡2xcos⁡3xdx=6sin⁡3x3∫0π−8⋅12[∫0πsin⁡x(sin⁡4x−sin⁡2x)dx]=6−82[∫0π(sin⁡xsin⁡4x−sin⁡x⋅sin⁡2x)dx]\begin{aligned} & =\int_0^\pi 2\left(3-4 \sin ^2 x\right) \cos 3 x d x \\ & =\int_0^\pi 6 \cos 3 x d x-\int_0^\pi 8 \sin ^2 x \cos 3 x d x \\ & =\frac{6 \sin 3 x}{3} \int_0^\pi-8 \cdot \frac{1}{2}\left[\int_0^\pi \sin x(\sin 4 x-\sin 2 x) d x\right] \\ & =6-\frac{8}{2}\left[\int_0^\pi(\sin x \sin 4 x-\sin x \cdot \sin 2 x) d x\right]\end{aligned}=−82[12∫0πcos⁡3x−cos⁡5x−cos⁡x+cos⁡3xdx]=−2[2sin⁡3x3−sin⁡5x5−sin⁡x]0π=0\begin{aligned} & =\frac{-8}{2}\left[\frac{1}{2} \int_0^\pi \cos 3 x-\cos 5 x-\cos x+\cos 3 x d x\right] \\ & =-2\left[\frac{2 \sin 3 x}{3}-\frac{\sin 5 x}{5}-\sin x\right]_0^\pi=0 \end{aligned}Hence,Hence,\mathrm{I}_2+I\mathrm{I}_3=0+0=0\therefore \quad$ Option (D) is correct.
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