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Quick integral question

  1. Apr 13, 2007 #1
    what is the quick way of doing single integrals of the form:

    *integral* (sinx)^n (cosx)^m dx

    where n and m are just integers. These kind of integrals come up all the time in vector calculus and they take me ages to do. Is there a general method of doing them or a few common integrals i could learn? I end up having to apply about 3 trig identites and then sorting out the mess of resulting sinxs, cosx, cos2x etc

    an eg would be

    *integral* cosx (sinx)^2 dx

    I would do this by using (sinx)^2 + (cosx)^2 = 1
    then cos^2= 1/2(1 +cos2A)
    then cosCcosD=1/2(cos(C+D) + cos(C-D))

    Which takes about a side and a half of a4. There must be a simpler way (substitutions, change of variables?)
  2. jcsd
  3. Apr 13, 2007 #2
    Most tables of integrals have a reduction formula for those kinds of integrals. But that's not exactly time saving either.
  4. Apr 13, 2007 #3
    This particular one is easy to do (using f'(x)dx=df(x) and substitution without a new named variable):
    [tex]\int \sin^n x \cos x dx = \int \sin^n x d(\sin x) = \frac1{n+1}\sin^{n+1}(x)[/tex]
    Last edited: Apr 13, 2007
  5. Apr 13, 2007 #4
    sorry im not sure i understand that. what is f'(x)dx=df(x)? (ive probably done that type of integration before but not with that name or notation. is that 'd' as in 'dx' or just a constant) and im not sure what substitution "without a new named variable means" either :confused:

    your method definately looks quicker and more sensible
  6. Apr 13, 2007 #5
    By df(x) he means the differential of f, which is equal to f'(x)dx
  7. Apr 13, 2007 #6
    It's the same as integration by substitution with u=sin(x). Work it out, and you'll see. I keep calling it sin(x) rather than u though, since I have to get back to it eventually.
  8. Apr 13, 2007 #7
    yeh, okay. and when you change the dx to dt you get 1/cosx which cancels the one in the integrand. very clever

    thankyou Eighty , much appreciated
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