anonymous
  • anonymous
if 0 < k
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
use equation option
anonymous
  • anonymous
\[\int\limits_{0}^{k} \cos(2x) dx =1/2\] thats what i meant
anonymous
  • anonymous
[sin(2k)/2]-sin(0/2)=sin(2k)/2=1/2 sin(2k)=1 2k=[\pi/2\] so k=\[\pi/4\]

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More answers

anonymous
  • anonymous
why do you divide by 2?
anonymous
  • anonymous
as you have 2x
anonymous
  • anonymous
in more detail put 2x=t..........(1) now differentiate both sides 2dx=dt dx=dt/2 and limits will change like this t=0 at x=0 .......from (1) t=2k
anonymous
  • anonymous
now solve interms of t
anonymous
  • anonymous
t is a random variable or what does it stand for?
anonymous
  • anonymous
random
anonymous
  • anonymous
sorry i dont get it still, i guess cause thats not how they have taught us over here...um are you showing me the antiderivative?
anonymous
  • anonymous
what's ur country name?
anonymous
  • anonymous
United States of America
anonymous
  • anonymous
k i was showing u the antiderivative
anonymous
  • anonymous
ohh ok.. i was told to add one to the power and divide by the new power, is that what you did?
anonymous
  • anonymous
it in the case of polynomial i mean sometjing like....x^2 or x^100
anonymous
  • anonymous
um ok?
anonymous
  • anonymous
if you have sin (polynomial of degreee 1) or cos(polynomial of degree 1) then simply do integretion 4 sin(x) or cos(x) then divide it by integretion of polynomial
anonymous
  • anonymous
thats why you divided by 2 since it was in the polynomial?
anonymous
  • anonymous
sorry its late and im a sort of slow now
anonymous
  • anonymous
ok add me as fan
anonymous
  • anonymous
lol
anonymous
  • anonymous
your last name is patel?
anonymous
  • anonymous
a friend of mine in calculus is last named patel he is from india, in any way related? i know big world but we found a cousin of his in a math competition lol
anonymous
  • anonymous
no i don't know him but i m also from INDIA
anonymous
  • anonymous
i guess its a popular last name from over there
anonymous
  • anonymous
yep
anonymous
  • anonymous
what's ur full name
anonymous
  • anonymous
mine?
anonymous
  • anonymous
yeah!
anonymous
  • anonymous
benjamin vasquez
anonymous
  • anonymous
got it any more q benjamin
anonymous
  • anonymous
nope! thank you so much

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