Differentiation of Integral

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  • #1
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Homework Statement




[tex]\frac{d}{dx}\int_{x^3}^{e^x}cost^2dt[/tex]



The Attempt at a Solution



[tex]\int cost^2dt=\frac{sint^2}{2t}+\int\frac{sint^2}{2t^2}dt[/tex]
[tex]\int\frac{sint^2}{2t^2}dt=-\frac{sint^2}{2t}+\int cost^2dt[/tex]
I came back to initial integral.
 

Answers and Replies

  • #2
Gib Z
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That can happen with some of your choice for u and dv whilst doing integration by parts, the second time you apply it use different choices.
 
  • #3
arildno
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Differentiate it, don't try to integrate it!
 
  • #4
Gib Z
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I think he needs to evaluate the integral to be able to do that doesn't he >.<
 
  • #5
arildno
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Nope.
Here's how to do it properly:
Let F(t) be an antiderivative of f, F'(t)=f(t).
Thus, we have:
[tex]\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}(F(b(x))-F(a(x)))=F'(b(x))b'(x)-F'(a(x))a'(x)=f(b(x))b'(x)-f(a(x))a'(x)[/tex]

As you can see, you do not need the explicit form of F, only the guarantee that some such F exists..:smile:
 
  • #6
D H
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I think he needs to evaluate the integral to be able to do that doesn't he >.<
No. The integrand does not involve x. Simply apply the fundamental theorem of calculus.

Hint:
[tex]
\frac{d}{dx}\int_{x^3}^{e^x}\cos t^2dt =
\frac{d}{dx}\int_0^{e^x}\cos t^2dt \;\;-\;\;
\frac{d}{dx}\int_0^{x^3}\cos t^2dt[/tex]
 
Last edited:
  • #7
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So tha answer is
[tex]2e^xsine^{2x}-6x^5sinx^6[/tex]
yes?
 
  • #8
D H
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No. Read arildno's post again. Sine is not involved.
 
  • #9
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ok ok,my mistake
[tex]2e^{2x}cos(e^{2x})-6x^5cos(x^6)[/tex]
 
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  • #10
arildno
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Eeh??
Where do you get that 2-factor from??
 
  • #11
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if we put [tex]e^x [/tex] to t shouldn't it be [tex]e^{2x}[/tex]
 
  • #12
arildno
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I'm talking about the 2-factors in front of the cosine's, not the ones within the arguments.
 
  • #13
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I typed wrongly instead of [tex]e^{2x}[/tex],I typed [tex]e^{x}[/tex]
[tex]\frac{d}{dx}(e^{2x})=2xe^{2x}[/tex]
This 2 are you asking ?
 
  • #14
arildno
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What is a(x), and what is b(x); what are their derivatives?
 
  • #15
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You say that answer is
[tex]e^xcos(e^{2x})-3x^2cos(x^6)[/tex]?
 
  • #16
Gib Z
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Today was not my best day obviously =] Yes I should have seen the proper method arildno and DH, maybe Ill have better luck tomorrow.
 
  • #17
HallsofIvy
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No, no one has said that! Several people have asked you questions about this problem that you haven't answered.
 
  • #18
arildno
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You say that answer is
[tex]e^xcos(e^{2x})-3x^2cos(x^6)[/tex]?

I didn't say that; but you said it correctly now! :smile:
 
  • #19
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Ok.Thanks to everyone.
 

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