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Substitution homework question

  1. May 26, 2006 #1
    Hi, Im having a little bit of trouble with the following:

    [tex]\int sintsec^2(cost)dt[/tex]

    heres what I have so far

    [tex]u=cost[/tex]

    [tex]du=-sintdt[/tex]

    [tex]-\int sec^2(u)du[/tex]

    [tex]-2tan(u) + C[/tex]

    is this right?
     
    Last edited: May 26, 2006
  2. jcsd
  3. May 26, 2006 #2
    As I recall the derivative of tan(u) is sec^2(u)du
     
  4. May 26, 2006 #3
    yea but im integrating..so the antiderivative of sec^2(u) is tanu?
     
  5. May 26, 2006 #4

    Hootenanny

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    [tex]\int \sec^2 x \; dx = \tan x +c[/tex]

    ~H
     
  6. May 26, 2006 #5
    I have another..i dont know where to start..can someone point out what I should sub U for?

    [tex]\int_{1/2}^{1/6}csc \pi t cot \pi t dt[/tex]
     
    Last edited: May 26, 2006
  7. May 26, 2006 #6

    Hootenanny

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    My first step would be to turn the cosec and cot into sine and cosine. See where that takes you.

    ~H
     
  8. May 26, 2006 #7
    ok..so i did..
    [tex] \int_{1/2}^{1/6} \frac{1}{sin\pi t) \frac{cos \pi t}{sin \pi t}dt[/tex]
    ended up with...
    [tex] \int_{1/2}^{1/6} cot \pi t[/tex]

    im guessing its wrong haha
     
  9. May 26, 2006 #8

    Hootenanny

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    It's almost there :wink:

    [tex]\frac{1}{\sin \pi t} \cdot \frac{\cos \pi t}{\sin \pi t} = \frac{\cos \pi t}{\sin^2 \pi t}[/tex]

    Now, see what you can do with the identity [itex]\sin^2 \theta + \cos^2 \theta = 1[/itex]. Btw, I think this can be done without a substitution.

    ~H
     
    Last edited: May 26, 2006
  10. May 26, 2006 #9
    do you mean..sin^2(pi t) = 1-cos^2(pi t)..and then sub?
     
  11. May 26, 2006 #10
    Forget about the subs.
    Try putting sin(pi*t) = x
    What is dx ?
     
  12. May 26, 2006 #11

    HallsofIvy

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    As others have pointed out, the anti-derivative of sec2(u) is tan(u), not -tan(u).

    Also, the original problem does not say anything about "u"! That was your "invention". To properly answer the question, you need to go back to t:

    [tex]\int sintsec^2(cost)dt= 2tan(cos(t)+ C[/tex]
     
  13. May 26, 2006 #12
    Actually, the original poster (with respect to the original question) is correct, save for the two.

    [tex] \int sin(t) sec (cos(x))^2 dx [/tex]

    [tex] u = cos(x) du = -sin(x) dx [/tex] This is right. Substituting back yields

    [tex] - \int sec(u)^2 du [/tex] Just as the OP said. The negative sign is the result of the du = -sin(x) dx part.

    Now, the antiderivative of sec(x)^2 = tan(x) + C (as everyone as stated)

    - ( tan(u) + C)
    -tan(cos(x) + C.

    The original poster's only error is the two infront. The minus sign is correct . Check with differentiatoin (or use the "Integrator").
     
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