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Various Questoins

  1. Jul 6, 2004 #1
    I just took a take home calc test this weekend, turned it in this morning. There were a few questions I couldn’t answer. Will you guys tell me (or even better show me) how to solve these?

    #1 [tex] \int \frac{dx}{3sinx - 4cosx} [/tex]

    #2 Find g’(x) where g is an inverse function of f(x)
    f(x) = 3 + x^2 + sin([pi]x) -0.4 < x < 0.4

    #3 find the exact value of sin[arcsine(1/3) + arcsine(2/3)]
  2. jcsd
  3. Jul 6, 2004 #2
    #1. You can rewrite 3sin(x) - 4cos(x) as k * sin(x + v), where k and v are constants. Then you just have integrate 1/k * csc(x + v).

    #2. Hmm, I wonder if you can leave g(x) in the answer?

    #3. Simpify using the addition formula for sine, and find a formula for cos(arcsin(x/y)).
    Last edited: Jul 6, 2004
  4. Jul 6, 2004 #3


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    Here is always a useful substitution for integrals of that and similar form:

    [tex]t = \tan \frac{x}{2}[/tex]

    [tex]\cos x = \frac{1-t^2}{1+t^2}[/tex]

    [tex]\sin x = \frac{2t}{1+t^2}[/tex]

    [tex]\frac{dx}{dt} = \frac{2}{1+t^2}[/tex]

    (Not 100% sure I have that last one right)
    Last edited: Jul 7, 2004
  5. Jul 6, 2004 #4
    #2 - f(x) = 3 + x^2 + sin([pi]x)
    g(x) is the inverse of f(x). Remember, just because it is written as g(x) doesnt mean that you have to make g a function of x. You are not asked to find the inverse of the function. You are rather being asked to find dy/dx of the inverse.

    The inverse of f(x) is:
    x = 3 + y^2 + sin([pi]y)

    now can you find dy/dx? (think of it as implicit differentiation)

    EDIT: nevermind you can't do this. I just realized they don't want dy/dx they instead want g'(x) which is not the same thing.
    Last edited: Jul 7, 2004
  6. Jul 6, 2004 #5
    How does this take-home test system work? What is stopping you from asking these same questions during the test period (other than your conscience - but who ever listens to it anyways?)?
  7. Jul 7, 2004 #6
    thats about it. It was 25 questions and these are the only 3 that i didn't get right i'm pretty sure.
  8. Jul 7, 2004 #7
    [tex]sin(x+y) = ?[/tex]
    [tex]sin(sin^-^1(x))= ?[/tex]
  9. Jul 7, 2004 #8
    forget to say it is just another way to solve the problem because the answer given by Muzza can also be applied..
  10. Jul 7, 2004 #9
    Seems to me like we were thinking of the same solution. Since you know, sin(x + y) will not only include sin(x) and sin(y), but also cos(x) and cos(y)...
  11. Jul 7, 2004 #10
    Yeah, i should have thought about what you wrote more deeper....Nothing big right ? --lol

    Uhmm, the same!
  12. Jul 7, 2004 #11
    Thanks for the help guys, those problems were a lot easier then I was making them out to be…

    i have a new question…

    Is the equation [tex] y = \lim_{n \rightarrow \infty} \pm(1 - x^{2n})^{1/2n} [/tex] a square?

    Is [tex] \lim_{n \rightarrow \infty} \int \pm(1 - x^{2n})^{1/2n} dx [/tex] = to 4? i.e. a 2x2 square?
    Last edited: Jul 7, 2004
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