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