1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mark44

    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

    Mark44

    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

    Mark44

    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

    so
    [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

    Mark44

    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

    Mark44

    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

    Mark44

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trigonometric Substitutions - help!
Loading...