Integral Calculus> options -nTT -(n +1) π -2nT -2(n +1) TT...

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

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 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.
$I = \int_{0}^{2n\pi }[sint]dt + \int_{2n\pi }^{(4n+1)\pi }[sint]dt$
$I = n\int_{0}^{2\pi }[sint]dt + \int_{2n\pi }^{4n\pi }[sint]dt + \int_{4n\pi }^{(4n+1)\pi }[sint]dt$
$[sint] = 0, 0\leq x\leq \pi/2$
$[sint] = 1,x = \pi/2$
$[sint] = 0, \pi /2\leq x\leq \pi$
$[sint] = -1, \pi \leq x\leq 2\pi$
$I = n(0-1)_{\pi }^{2\pi} + \int_{2n\pi }^{4n\pi }[sint]dt$
$I = -n\pi + \int_{2n\pi }^{4n\pi }[sint]dt$
$I = -n\pi + (2n-n)\int_{0 }^{2\pi }[sint]dt$
$I = -n\pi + n\int_{\pi }^{2\pi }(-1)dt$
$I = -2n\pi$