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Antiderivative of product of trig functions

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the antiderivative of [tex]sin^4(x)tan^2(x)[/tex]


    2. Relevant equations
    Trig identities I may have overlooked.

    3. The attempt at a solution
    I tried writing the integrand in terms of sin and cos but that didn't seem to lead anywhere. I tried integration by parts since the antiderivative of [tex]tan^2(x)[/tex] is [tex]tanx - x + C[/tex], but that approach lead to a more difficult integral that might have been solved using a messier integration by parts.

    This isn't really homework since I'm in a theoretical calc course and we begin the theory of integration next quarter (in fact, we're on break). Anyways I used to be pretty good with elementary terms integration but I'm stuck. Could someone suggest a good first step? Thanks.
     
  2. jcsd
  3. Dec 11, 2008 #2

    Mark44

    Staff: Mentor

    I think this will go with integration by parts.
    Try u = tan^2(x)sin^3(x) and dv = sin(x)dx.

    This gave me an unwieldy-looking integral, but after simplifying to sines and cosines, I was left with [tex]\int \frac{sin x}{cos^2 x} dx[/tex] and [tex]\int sin^4 x dx[/tex]. There were constant multipliers that I have omitted.

    The first integral is an easy one and the second isn't too hard.
    Hope that helps.
     
  4. Dec 11, 2008 #3
    Another potential way to do this integral is by converting everything to cosines because there is a cosine downstairs:

    [tex]\int \sin^4x\,\tan^2x\,dx = \int\frac{(1-\cos^2 x)^3}{\cos^2 x}\,dx = \int \frac{1 - 3\cos^2 x + 3\cos^4 x - \cos^6 x}{\cos^2 x}\,dx[/tex]​

    Each of the resulting integrals should be fairly simple to solve.
     
  5. Dec 11, 2008 #4
    Thank you both for the suggestions.

    Mark44, did you arrive at those two integrals after applying integration by parts once? I got sin sin^4(x) term but the the one before that (I think it was sin^4(x)/cos^2(x) or something). I'll try it again though.

    Tedjn, I thought of the very first equality you posed but did not think to expand it. That is quite a neat trick, i.e., decrease the powers by expansion. A few expansions and some computation and it worked out well. Thanks.
     
  6. Dec 12, 2008 #5

    Mark44

    Staff: Mentor

    Yes.
     
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