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Simple? Integration Problem (Trig Sub?)

  1. Feb 4, 2008 #1
    I can't get it! I'm pretty sure it's trig substition


    Its a practice problem, if someone could show me the light (or steps) that would be wonderful
  2. jcsd
  3. Feb 4, 2008 #2
    well, yeah a trig substitution would work, try to let x=sin(t) , so you will get dx=cos(t)dt
    after you substitute it back you willl end up with something like this

    integ of (sin(t))^2dt

    Is this a homework problem by the way???

    Can you go from here, anyway??
  4. Feb 4, 2008 #3
    No its not a homework problem....its a conceptual problem ....but i dont get it, and I'm not that great with calculus to tell you the truth im not a math major i just want to get it! So is this what you mean....

    [tex]\int(sinx)^{2}[/tex][tex] / [/tex][tex]\sqrt{1-sinx^{2}}[/tex]

    so using u subtitution ( or whatever letter you use)...
    du= cosx
    and since there is no cos in the original then 1/cos(du)

    (1/cos)[tex]\int du(u)^{2}[/tex][tex] / [/tex][tex]\sqrt{1-u^{2}}[/tex]

    and go from there? did i do it right?
  5. Feb 4, 2008 #4
    Well You did not get it right, to be honest. Look, [tex]\int\frac{x^{2}}{\sqrt{1-x^{2}}}dx[/tex] now let sin(t)=x, from here after defferentiating we get cos(t)dt=dx, now let us substitute this back to the integral, so the integral will take this form:
    [tex]\int\frac{(sin(t))^{2}}{\sqrt{1-(sin(t))^{2}}}cos(t)dt[/tex], now remember that (sin(t))^2= 1-(cos(t))^2, so afer we substitute the integral becomes:
    [tex]\int\frac{(sin(t))^{2}}{\sqrt{(cos(t))^{2}}}cos(t)dt[/tex]= [tex]\int\frac{(sin(t))^{2}}{cos(t)}cos(t)dt[/tex]= [tex]\int (sin(t))^{2}dt[/tex], now do u know how to evaluate this one?
    Last edited: Feb 4, 2008
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