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Trigonometric Integration, Definite integral.

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Use 2 or more substitutions to find the following integrals
    hint : begin with u=cosx


    2. Relevant equations
    Integral 0--->pi/2 (cosxsinx)/swrt(cos2x+49 dx


    3. The attempt at a solution
    I'm still a little fuzzy on using multiple substitutions. From what I've read on the text and previous easier equations, it just means that there are multiple u=(something) that can work. Is that right?

    so I tried u=cosx
    du=-sinxdx

    giving me


    -1 * Integral 0--->pi/2 u/sqrt(u2+49) du


    it's here that I am brickwalling. I really want to know how it works, so if you wouldn't mind a step by step process, I'd appreciate greatly.
     
  2. jcsd
  3. Jul 24, 2011 #2

    SammyS

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    You should either change the limits of integration, or do the corresponding indefinite integration.

    You have [itex]\displaystyle \int \frac{u}{\sqrt{u^2+49}} du[/itex].

    Can you see a substitution which might work with your result?

    (I can see two, either of which looks helpful.)
     
  4. Jul 24, 2011 #3
    So I beat at it until I solved it :D ( I dont give up dammit)

    I put the integral into terms of u
    so
    u=cos(0)=1
    u=cos(pi/2)=0


    so

    integral 0-->1 u(u^2+49)^-1/2 du
    I took the second sub of t=u^2+49
    dt=2udu
    to give
    1/2 integral0-->1 (t)^-1/2 dt
    1/2 * 2 (t)^1/2
    giving the function
    (u^2+49)^1/2 |0-->1
    then using the fundamental theorem of calculus
    [(1^2+49)^1/2]-[0^2+49]^1/2]
    sqrt(50)-sqrt(49)
    giving
    -7+5sqrt(2)

    :D

    pretty stoked lol
     
  5. Jul 24, 2011 #4

    SammyS

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    Excellent! (and welcome to PF !)

    Another (very nice) subst. would have been to let t = (u2+49) .

    Try it, you might like it.
     
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