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How can I do integrals by just eyeing them?

  1. Dec 11, 2009 #1
    So I have an integral say:

    [tex]
    \int \! \cos 3x dx[/tex]

    The antiderivative is
    [tex]\frac{1}{3}\sin (3x) + K[/tex]

    But it's not obvious to me how to eye it. What I normally do is try to look for a derivative of something within the integral, so I'd know it's a result of a chain rule, but this function doesn't have it. Then I try integration by parts and it just becomes a whole big mess. I also tried the opposite rule of differentiation where you just raise the exponent by 1, then divide by the exponent, which doen't work in this case since it's not an easy polynomial.

    Finally, I'm out of options and I do guessing and checking by doing semi random functions and differentiating them to see if I get the integral, and if I do, I've found the antiderivative.

    There's got to be a better way to interpret this problem. I think I'm just not thinking of it correctly. Can anyone give me some pointers?
     
    Last edited: Dec 11, 2009
  2. jcsd
  3. Dec 11, 2009 #2
    For me, being able to "eye" an integral comes with practice. There are still many integrals that are harder to see the answer right away.

    With your function, if y = 3x then dy = 3dx. In order to get 3dx inside of the integral, it needs to be balanced by a 1/3, which can be brought outside of the integrand:

    [tex]\frac{1}{3}\int\cos(3x)3dx[/tex]

    [tex]\frac{1}{3}\int\cos(y)dy = \frac{1}{3}\sin(y) + K = \frac{1}{3}\sin(3x) + K[/tex]

    Always good to check:

    [tex]\dfrac{d}{dx}(\frac{1}{3}\sin(3x) + K) = \frac{1}{3}\cos(3x)\dfrac{d}{dx}3x = \frac{3}{3}\cos(3x) = \cos(3x)[/tex]
     
  4. Dec 11, 2009 #3
    Maybe I'm looking too much into it when this should be a more basic question.

    What does the dx (or d(whatever)) mean? Is it really necessary to have the dx beside every integral? I always forget to leave that in for some reason, and questions still seem to work out.
    d/dx = derivative with respect to x, dx = ?
     
  5. Dec 12, 2009 #4
    Yes it is necessary for your expression to make sense. It is there because (in an intuitive way) you ARE multiplying the integrand by dx. You evaluate the integrand at the starting point, multiply it by dx, then scoot along in the functions variable by an amount dx to f(x+dx), then multiply that function value by dx, and so on then sum all of them. The notation of the integral contains all of that information in there. It literally means "take the sum of this infinite number of products of function values with dx". That is why the thing is a squiggly line, the integral sign is just a stylized S, meaning take the sum of...
     
  6. Dec 12, 2009 #5
    I'm understanding what integrals are, they're basically summation notation of the area in a curve, or something like that... I was just having trouble with notation. Is Pbandjay's solution to the problem considered u-substitution? I thought you only use u-substitution when you see an expression and its derivative within an integral.
     
  7. Dec 12, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    For a linear substitution, u= ax+ b, du= adx or dx= (1/a)du. You can always move constants, like 1/a, in or out of integrals. For anything more complicated, du will be some function of x times dx and that cannot be moved in or out of the integral- it must already be "within" the integral.

    "Substitution" in an integral is basically the inverse of the chain rule. If you see something like f(ax+b), you can, with practice, think immediately that its derivative is af'(ax+b). For f(u(x)) where u is more complicated than just ax+b, you might have to work out the derivative of u. Similarly, with [itex]\int f(ax+b) dx[/itex] you can, with practice, think immediately that it is (1/a)F(ax+b)+ C where "F" is an anti-derivative of f. For [itex]\int F(u(x))dx[/itex] where u is more complicated than just ax+ b, that is a much more difficult problem.
     
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