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Haven't been taught integration by parts yet

  1. Jan 15, 2009 #1
    [​IMG]


    Attempt to the solution. I took u=[tex]\sqrt{t}[/tex] and from there i went upto :

    2[tex]\int[/tex]u[tex]^2[/tex]Sin(u^{2})dx

    dunno wat to do next. Havn't been taught integration by parts yet.
     
  2. jcsd
  3. Jan 15, 2009 #2
    Re: Integral!

    What is the question? Are you sure it isn't to find g'(x)?
     
  4. Jan 15, 2009 #3
    Re: Integral!

    yes it is.. sorry my bad. :P

    then we have to expand the integral and and simply write the inner function by replacing t with x rite?

    (i spend much time figuring out the integral, but in vain.. :P)
     
  5. Jan 15, 2009 #4

    Mark44

    Staff: Mentor

    Re: Integral!

    Your substitution is not right. If u = [itex]\sqrt{t}[/itex], then du = dt/(2[itex]\sqrt{t}[/itex])
    dx shouldn't even be in the integral.
     
  6. Jan 15, 2009 #5
    Re: Integral!

    If you are interested to find g'(x), then you don't even need to do any integration at all. All, you need to do is apply 1st part of the FUndamental THeorem of Calculus. What does it say?

    First manipulate your integral a little bit, like:

    [tex]g(x)=-\int_{c}^{\sqrt{x}}\sqrt{t}sin(t)dt+\int_{c}^{x^2}\sqrt{t}sin(t)dt[/tex]

    where c is some constant.

    Now you need to directly apply FTC and the chain rule along with it and you are pretty much done.
     
  7. Jan 15, 2009 #6
    Re: Integral!

    Apart from the dx (which I corrected in the quotation in my response), the substitution, ignoring the limits, is right. But as Sutupidmath has correctly followed up, this is a futile approach if one wants to find g'(x) anyway!
     
  8. Jan 15, 2009 #7

    Mark44

    Staff: Mentor

    Re: Integral!

    My mistake.
     
  9. Jan 15, 2009 #8
    Re: Integral!

    If you need to find g'(x) then remember this

    [tex] f(x) = \int_{\alpha(x)}^{\beta(x)} \zeta(t) dt \Rightarrow f'(x) = \zeta(\beta(x)) \cdot \beta'(x) - \zeta(\alpha(x)) \cdot \alpha'(x) [/tex]

    Hope that helps.
     
  10. Jan 18, 2009 #9
    Re: Integral!

    Thanks alot. I got it now !! :) A very useful rule!
     
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