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2 integration problems

  1. Apr 15, 2009 #1
    These aren't really homework, but they're probably easy enough to be... so =). These are off some sheet that I found and decided to do, and have gotten stuck on various ones.

    1. Question 1

    2. [tex]\int \frac{\mbox{dx}}{x\sqrt{x^2-a^2}}[/tex]

    3. The attempt at a solution

    Okay, first of all, there have been 2 questions prior to this one which only differ under the square root sign; the [tex]a^2[/tex] has come before the [tex]x^2[/tex] in both cases and could be done with my crappy knowledge of doing a trig substitution. However this one has gotten me confused; the [tex]x^2[/tex] term is now before the [tex]a^2[/tex] and I don't think I'm finding the right trig substitution. The sheet provides an answer of [tex]\frac{1}{a}\ arcsec \frac{x}{a}[/tex]... and I have no idea how that came about.


    Question 2 to come next.
     
    Last edited: Apr 15, 2009
  2. jcsd
  3. Apr 15, 2009 #2

    Cyosis

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    Hint, [tex] \tan^2 x+1=\sec^2x[/tex].
     
  4. Apr 15, 2009 #3
    1. Question 2

    2. [tex]\int \frac{dx}{1+cos^2 x}[/tex]

    3. An attempt at a solution

    I've tried doing various u-substitutions which went nowhere, I've tried making trig identities, but I can't seem to find one that works. I've also tried splitting the 1 dx on top into [tex]sin^2 x+cos^2 x[/tex] etc. but have gotten nothing...
     
  5. Apr 15, 2009 #4
    Okay let's see:

    let [tex]x=asec \theta[/tex]
    then [tex]dx=atan \theta sec \theta[/tex]

    Edit: That comes to give [tex]\int \frac{1}{a} d\theta[/tex]
    That looks right.
     
    Last edited: Apr 15, 2009
  6. Apr 15, 2009 #5

    Cyosis

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    That's because you need to make the substitution [itex]x=a \sec \theta[/itex] and not [itex]x=a \sec^2 \theta[/itex]. With your substitution you get a [itex]\sec^4 \theta[/itex] so the entire rule doesn't apply.

    Hint 2 (for integral 2): [tex]\cos 2x=2 \cos^2 x-1[/tex].
     
  7. Apr 15, 2009 #6
    Oops, that's what I meant; not squared... better edit that.
     
  8. Apr 15, 2009 #7

    Cyosis

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    Well if that's what you meant you should note that your dx is wrong as well, which is the reason you don't get the right answer.

    I see you edited your post with a question. If you get [itex]\frac{1}{a} \theta[/itex], what is [itex]\theta[/itex] in terms of x?
     
    Last edited: Apr 15, 2009
  9. Apr 15, 2009 #8
    Now that I've edited the above...

    it becomes
    [tex]\frac {1}{a} \theta[/tex]
    and [tex]\theta = arcsec \frac{x}{a}[/tex]

    Yes, that looks right now. Thanks.


    Edit: However with your second hint, I am still struggling :(
     
    Last edited: Apr 15, 2009
  10. Apr 15, 2009 #9

    Cyosis

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    What have you tried so far with the second hint?

    Ugh, I misread your second integral overlooked the square, so that hint is not going to work, apologies.

    This doesn't seem to be an easy one but the following method works.

    Some intermediate steps:
    first substitute [tex]x=\arccos u \Rightarrow dx=\frac{-1}{\sqrt{1-u^2}} du[/tex]. Then substitute [tex]u=\frac{1}{\sqrt{1+z^2}} \Rightarrow du=\frac{-z}{(1+z^2)^{\frac{3}{2}}}dz [/tex].

    If you've done everything correctly you get to the following integral:

    [tex]\int \frac{dz}{2+z^2}[/tex]

    You then cast this integral into a form where you recognize the derivative of the arctangent.
     
    Last edited: Apr 16, 2009
  11. Apr 15, 2009 #10
    Thanks for staying with me here Cyosis;

    I understand both the substitutions you have provided (by that I mean I know what you're talking about in terms of d/dx etc) but I don't know how I can use the second one.
    Or is it a straight substitution after the first one you have provided?
     
  12. Apr 16, 2009 #11

    Cyosis

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    Just plug them in one after another. It will look like a mess but you can simplify it quite easily to the simple form in my previous post. Alternatively you could also use u=tan x, this way only requires one substitution.
     
    Last edited: Apr 16, 2009
  13. Apr 16, 2009 #12
    I've done as you've guided me

    And now I've got (after canceling etc etc)

    [tex]\int \frac {dz}{(z^2+2)(z^2+1)^\frac {1}{2}} [/tex]

    What have I done wrong now...?

    I've got the [tex]z^2+2[/tex] but I have a leftover [tex](z^2+1)^\frac {1}{2}[/tex] from the du/dz having the index of 3/2...
     
  14. Apr 17, 2009 #13

    Cyosis

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    After doing the double substitution you should have gotten this integral. I reckon you made a mistake with the root, but I can't tell from that expression.

    [tex]
    \int\frac{1}{(\frac{1}{1+z^2}+1) \sqrt{1 - \frac{1}{1+z^2}}} \frac{z}{(1+z^2)^{\frac{3}{2}}}\,dz
    [/tex]

    Hint:[tex](1+z^2)^{\frac{3}{2}}=(1+z^2) \sqrt{1+z^2}[/tex].
     
  15. Apr 18, 2009 #14
    Yeah I think I got that

    so if you multiply top and bottom by [tex]z^2+1[/tex]...

    I get

    [tex]\frac {z(z^2+1)}{(z^2+2)\sqrt{z^2}(z^2+1)^\frac {3}{2}}[/tex]

    so everything cancels and I had the root 1+z^2 leftover (incl the z^2 + 2) in the bottom... or am I just doing something plain stupid here?

    However, I thought back to what you said very early about the [tex]cos^2 x = \frac {1}{2}(1+cos2x)[/tex] and used it...

    Then I did some working and used t formula

    which eventually got me to [tex]\int \frac {dt}{2+t^2}[/tex] from which I was able to get the answer.

    But regardless, thanks for all your help. Maths really can be !%A&
     
  16. Apr 18, 2009 #15

    Cyosis

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    It seems you took somewhat of a detour, but the correct answer is the correct answer regardless. The following was my intention however:

    [tex]
    \begin{align*}
    \int\frac{1}{(\frac{1}{1+z^2}+1) \sqrt{1 - \frac{1}{1+z^2}}} \frac{z}{(1+z^2)^{\frac{3}{2}}}\,dz &= \int\frac{1}{(\frac{1}{1+z^2}+1) \sqrt{1 - \frac{1}{1+z^2}}} \frac{z}{(1+z^2) \sqrt{1+z^2}}\,dz
    \\
    &= \int\frac{1}{(\frac{1}{1+z^2}+1)(1+z^2) \sqrt{1 - \frac{1}{1+z^2}}} \frac{z}{\sqrt{1+z^2}}\,dz
    \\
    &= \int\frac{1}{(1+1+z^2) \sqrt{1+z^2-1}}z \,dz
    \\
    &=\int\frac{z}{(2+z^2)z} \,dz
    \end{align*}
    [/tex]
     
  17. Apr 18, 2009 #16
    Ah so what I did do was stupid; made extra work for myself which indeed gave me the wrong answer. Thanks for the explanation.
     
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