anonymous
  • anonymous
why integral of( sin x )=-(cos x) ? and how?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
What is the derivative of cosx??
anonymous
  • anonymous
- sin x
anonymous
  • anonymous
Or you can say that the derivative of -cosx is sinx... Now integral is just reverse of derivative.. So whose sinx is derivative of which quantity???

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anonymous
  • anonymous
|dw:1342100312222:dw|
anonymous
  • anonymous
Or you can use Euler's Identities to prove it...
anonymous
  • anonymous
\[\large \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}\] \[\large \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}\] Find the integral of sin theta you will get -cos theta..
anonymous
  • anonymous
Other possibilities are to integrate the Taylor Series for Sin x on some interval or from first principles as here http://www.math.com/tables/derivatives/more/trig.htm
lgbasallote
  • lgbasallote
a good thing to remember is the integral of a cofunction is alsways negative that explains the negative sign
across
  • across
lmao she has no clue how to integrate \(\sin\) and there's people suggesting she use Euler's identity, hahaha.
anonymous
  • anonymous
thanks to every one i got the ans!!!!!!
anonymous
  • anonymous
She has clue how to integrate but she is asking why?? and mind your language @across
across
  • across
@DHASHNI, this follows from the differentiation of the trigonometric functions:\[\sin x\implies\cos x\\\cos x\implies-\sin x\\-\sin x\implies-\cos x\\-\cos x\implies\sin x\\\]
anonymous
  • anonymous
@across prove them...
across
  • across
I can Google you a proof in less than five seconds. I don't have to prove anything to you.
anonymous
  • anonymous
That is what she is asking... Google??? What else you can do...
anonymous
  • anonymous
@across : i knw the formulas ....... i jus wanna know how integral sinx is (-cos x)........the proof for that~
lgbasallote
  • lgbasallote
here's another fun proof \[\int sinx dx\] let u = cosx du = -sin x dx -du = sinxdx \[\implies\int -du\] \[\implies -\int du\] \[\implies -u\] \[\implies -\sin x\]
anonymous
  • anonymous
Now did you get what she said @across .. If you don't know how to prove them then do not make fun of anybody...
across
  • across
OH, she wants the PROOF. For all we know, the title may have suggested a geometric interpretation.
anonymous
  • anonymous
I think a word known as WHY is sufficient for all the things...
anonymous
  • anonymous
LOL

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