f(x)=2sinxcosx, what is the primitive function?
you have to find f'(x)
That would be f'(x)=2cos^2(x)-2sin^2(x)
I believe atleast, having such a hard time following productrules et cetera backwards.. :x
\[\Large f'(x)=(2 \cdot \sin x \cdot \cos x)'\] take number 2 off... \[\Large f'(x)=2( \sin x \cdot \cos x)'\] and go on...
http://www.sosmath.com/CBB/viewtopic.php?f=3&t=57694 the same problem I had before !
I think you're supposed to rewrite the function into something more manageable.
Perhaps by trigonometric identities or w/e, bah - clueless here!
I don't understand you O_O
is in the link what you're asking for or not ? ...(Google translate sucks !)
Sorry, English is not my native language and I take math in Sweden - quite hard to "translate" the math into English, hehe.
I cant put the thing you linked into context, hmm
Primitive means antiderivative so integrate it. \[\int\limits2sinxcosx \] let u = cosx => du/-sinx
@Mimi_x3 but how? ... then my task there was wrong , wan't it ? O_O
I'm aware of that Mimi_x3 The thing is - I don't know how to do it! :D
I remember when primitive means to integrate..
not differentiate.
Indeed it is the antiderivative.
aufff... then I apologize , I learned it wrong :( ... thanks for explaining it @Mimi_x3
Well, it's a u-substitution.
Hm, but I'm not sure, lol.
You're supposed to do it without substitution I believe, as the book I have explains substitution in the upcoming chapter.
You can work backwards..when you find the derivative i think..
I think what you basicly should do is use the product rule backwards, or rewrite it with the help of some trig identity.
Indeed, it's very tiresome and confusing though, haha :(
.. if it's integral ... then check this out ! ... http://www.wolframalpha.com/input/?i=2sinxcosx
But she has not learnt u-substition.
PRIMITIVE = INTEGRAAAAAALLLL!!!!!
I'm out !
write that as sin2x. then integrate using std. formula!!! should get ----> (-cos2x)/2 + constant
I shall try, thanks everyone!
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