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[Help] Integral Problem

  1. Dec 13, 2006 #1
    Ok I cant figure out how to solve this. I should be able to just plug in the sin(x) into the t's but it says that is the wrong answer. I asked my prof in class today to solve it, but he really didnt help. So maybe you guys can help me out. Thanks in advanced.

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  2. jcsd
  3. Dec 13, 2006 #2


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    Homework Helper

    What does the fundamental theorem of calculus say? Also, you'll need to use the chain rule.
  4. Dec 13, 2006 #3
    Its says the derivative of g(x) is equal f(x), I know the Fundamental Theorem of Calculus, but I just dont know how to use it alongside the chain rule. Our book only has one example of it, and it doesnt explain it very well.

    I think the answer is 5x^4*cos(x^5)*cos(sin(x)^5)+sin(x) but, I know its wrong cus the website wont take it.
    Last edited: Dec 13, 2006
  5. Dec 13, 2006 #4


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    Yea, but that's meaningless unless you define f(x) and g(x). I'm sure you do know it, but all this problem consists of is carefully applying the definitions, so you should be explicit.
  6. Dec 13, 2006 #5


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    No matter what f and g are? What a remarkable theorem!:rofl:

    One form of the fundamental theorem of calculus says that if
    [tex]g(x)= \int_{x_0}^x f(t)dt[/tex]
    then the derivative of g(x) if f(x). Here your only "problem" is that the upper limit is sin(x) instead of x. Can you make a substitution to correct that? As StatusX says, use the chain rule.
  7. Dec 13, 2006 #6
    Alright that didnt help me at all. Um, lets see can you just tell me the answer?!?! I think I would be able to see what Im doing wrong if I knew the answer.
  8. Dec 13, 2006 #7


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    No. Keep trying, or ask more questions.
  9. Dec 13, 2006 #8
    K the answer is cosx*cos(sin(x)^5)+cosx*sinx
  10. Dec 13, 2006 #9
    An awesome way to think of it is to say,

    [tex]h(x)=\int_{-5}^{\sin x} f(t) dt = F(\sin x) - F(-5)[/tex]

    You want d/dx of that, so use the chain rule straight off...

    [tex] \dfrac{d}{dx}(F(\sin x)) = \cos x (f(\sin x))[/tex]

    Where [tex]f(t) = \cos (t^5) + t[/tex]
  11. Dec 14, 2006 #10
    heh - just found a great site for this sort of thing the other day. I can't help you but these guys have sure helped me - great physics tutors cheap plus old q & a even cheaper from their library - great for studying.

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