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Math Help - calculate the derivative of an integral

  1. #1
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    calculate the derivative of an integral

    d/dx \int_x^0 \! sin^2(t) \, \mathrm{d}x.

    so first I flipped it & changed the sign to get the x on top:
    d/dx - \int_0^x \! sin^2(t) \, \mathrm{d}x.

    Now my question is, what is the integral of sin^2(t)?

    Like I know for just integral sinxdx=-cosx...

    If I set cos2x=1-sin^2(x)..will this work?
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  2. #2
    Super Member Aryth's Avatar
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    First, recognize that a function of t is independent of x and is therefore a constant with respect to x, this turns the integral into:

    sin^2(t)\frac{d}{dx} \int_x^0 ~dx

    Now, solve the integral, which is just x evaluated from 0 to x:

    sin^2(t)\frac{d}{dx} x|^0_x

    sin^2(t)\frac{d}{dx} -x

    Now we take the derivative, which is just -1:

    \frac{d}{dx}\int_x^0 sin^2(t) ~dx = -sin^2(t)
    Last edited by Aryth; January 23rd 2011 at 08:02 AM. Reason: Added some explanation
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  3. #3
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    But if you meant \frac{d}{dx}\int_x^0 sin^2(t)dt= -\frac{d}{dx}\int_0^x sin^2(t)dt ("dt", not "dx")
    then you could integrate by using the trig identity " sin^2(t)= \frac{1}{2}(1- cos(2t))".

    However integrating at all misses the point of this problem- it is an exercise in the "Fundamental Theorem of Calculus":
    \frac{d}{dx}\int_a^x f(t)dt= f(x).

    Do it both ways as a check!
    Last edited by HallsofIvy; January 23rd 2011 at 09:06 AM.
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  4. #4
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    ok, thank you so much! that made so much more sense than my book!
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