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Homework Help: Trigonometric Substitutions - help!

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]$\int \frac{\sqrt{1-4x^2}dx}{x}$[/tex]

    2. Relevant equations

    3. The attempt at a solution
    im stuck and i have no idea why im getting the wrong answer.

    let 2x = sin[tex]\phi[/tex]
    dx = cos[tex]\phi[/tex] d[tex]\phi[/tex] / 2
    [tex]$\int \frac{\sqrt{1-(2x)^2}dx}{x}$[/tex]=[tex]$\int \frac{\sqrt{1-(sin\phi)^2}dx}{x}$[/tex]=[tex]$\int \frac{\sqrt{cos\phi^2}dx}{x}$[/tex]=[tex]$\int \frac{cos\phi dx}{x}$[/tex]=

    [tex]$\int \frac{cos\phi^2 d\phi}{2x}$[/tex]= [tex]$ .5\int \frac{cos\phi^2 d\phi}{x}$ =[/tex] [tex]$ .5\int \frac{.5(1+cos(2\phi)) d\phi}{x}$= [/tex]
    [tex]$ (1/4) \int 1/x+cos(2\phi)/x d\phi $[/tex]

    am i even on the right truck?
  2. jcsd
  3. Feb 15, 2010 #2


    Staff: Mentor

    When you do your trig substitution, replace every x in your original integral. x is still appearing in the denominator.
  4. Feb 15, 2010 #3
    [tex] $\int \frac{cos\phi dx}{2sin\phi}$ = [/tex]
    [tex] $.5 \int \frac{cos\phi dx}{sin\phi}$ =[/tex]
    [tex] $.5 lnsin\phi + C$[/tex]

    but i dont think that's right is it..
  5. Feb 15, 2010 #4


    Staff: Mentor

    Nope. Start again from the beginning, and exchange x and dx for phi and dphi all in one step.
  6. Feb 15, 2010 #5
    so i did the same thing over again and i got this:
    [tex]$ (1/4) \int 1/sin\phi - sin\phi d\phi $ [/tex]
    which =
    [tex]$ (1/4)[ \int 1/sin\phi d\phi + cos\phi]$ [/tex]
    right? now what..
  7. Feb 15, 2010 #6


    Staff: Mentor

    dx = (1/2)cos t dt, and x = (1/2) sint -- t being short for theta

    The dx in the numerator and the x in the denominator cause the (1/2)'s to cancel.

    So you should have
    [tex]\int \left[\frac{1}{sin t} - sin t\right]dt[/tex]
    [tex]=\int \frac{1}{sin t}dt - \int sin t dt[/tex]

    For the first you need the antiderivative of csc t. The second is easy.

    Does you book show you a trick for antidifferentiating sec t? It involves multiplying by 1 in the form of sec t + tan t over itself. There's a similar trick for working with the integral of csc t.
  8. Feb 23, 2010 #7
    olympics distracted me for a while lol

    yeah i got that antiderivative after googling it hhh.. my book didnt say anything about it.

    now im stuck on this question:
    [tex]$\int \frac{dx}{\sqrt{16-x^2}}$[/tex]

    i let x = 4sint
    dx= 4cost dt

    [tex]$\int \frac{dx}{\sqrt{16-16sin^2}}$[/tex] =[tex]$\int \frac{4cost dt}{\sqrt{16(1-sin^2 t)}}$[/tex] =[tex]$\int \frac{4cost dt}{\sqrt{16cos^2 t}}$[/tex]=[tex]$\int \frac{4cost dt}{4cost}$[/tex] =[tex]$\int dt$[/tex]

    i dont see why im getting a total wrong answer :S
  9. Feb 24, 2010 #8


    Staff: Mentor

    = t + C

    Now undo your substitution. You had x = 4sint ==> x/4 = sint ==> t = ?
  10. Feb 24, 2010 #9
    oh thats what it is. so the answer would be [tex]$\arcsin (x/4)+C$[/tex]

    now my problems get more complicated, [tex]$\int \sqrt{5+4x-x^2}dx$ [/tex]
    my textbook shows a simpler question and it works there but this question is a bit different.
    i know that the first step is putting it like this: [tex]$\int \sqrt{5+(4x-x^2)}dx$ [/tex]
    then something has to come out and then the substitution comes in..
    whats my next step though?
  11. Feb 24, 2010 #10


    Staff: Mentor

    Write the part in the radical as 5 - (x^2 - 4x). Then complete the square so that you get something that looks like a - u^2 in the radical. Use a trig substitution on that.
  12. Feb 24, 2010 #11
    i was thinking about this 5 - (x2 - 4x) and i was thinking, if i let y2 = x then
    5 - (x2 - (2y)2)

    would that work? because now im on x2−a2
    im not sure though
  13. Feb 25, 2010 #12


    Staff: Mentor

    I don't think that would work. Try what I suggested. If you had sqrt{3 + 2x - x^2)}, you would do this:
    sqrt{3 + 2x - x^2)} = sqrt{3 - (x^2 - 2x)}
    = sqrt{3 - (x^2 - 2x + 1) + 1} = sqrt{4 - (x - 1)^2}

    If you let u = x - 1, you have sqrt{4 - u^2} and you're ready to do a trig substitution.

    For my example, draw a right triangle with the hypotenuse labeled 2, the opposite side labeled u, and the adjacent side labeled sqrt{4 - u^2}. One acute angle would be labeled theta or t or whatever (but not x or u, since those are already in use).

    Do the same on your problem.
  14. Feb 25, 2010 #13
    thats the exact same example from my textbook..
    so what i tried is, 5 - (x^2 - 4x +1) + 1
    = 6 - (x^2 - 4x +1)
    how do you make it ^2?
  15. Feb 27, 2010 #14
    okay so i got
    9 - (x-2)^2

    now i substitute u=x-2, du=dx
    i get [tex] $\int \sqrt{9 - u^2}du$ [/tex]

    then u=3sint
    du = 3cost dt
    [tex] $\sqrt{5+4x-x^2}dx$[/tex] = 3cost
    so [tex] $\int 3costdu$ [/tex]

    am i right so far?
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