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options -nTT -(n +1) π -2nT -2(n +1) TT

options
  1. -nTT
  2. -(n +1) π
  3. -2nT
  4.    
    -2(n +1) TT

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{0}^{x}[sint]dt
x\in (2n\pi ,(4n+1)\pi )
I = \int_{0}^{2n\pi }[sint]dt + \int_{2n\pi }^{(4n+1)\pi }[sint]dt
[sint] is a periodi with a period of 2pi.
I = \int_{0}^{2n\pi }[sint]dt + \int_{2n\pi }^{(4n+1)\pi }[sint]dt
I = n\int_{0}^{2\pi }[sint]dt + \int_{2n\pi }^{4n\pi }[sint]dt + \int_{4n\pi }^{(4n+1)\pi }[sint]dt
[sint] = 0, 0\leq x\leq \pi/2
[sint] = 1,x = \pi/2
[sint] = 0, \pi /2\leq x\leq \pi
[sint] = -1, \pi \leq x\leq 2\pi
I = n(0-1)_{\pi }^{2\pi} + \int_{2n\pi }^{4n\pi }[sint]dt
I = -n\pi + \int_{2n\pi }^{4n\pi }[sint]dt
I = -n\pi + (2n-n)\int_{0 }^{2\pi }[sint]dt
I = -n\pi + n\int_{\pi }^{2\pi }(-1)dt
I = -2n\pi





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