integral question - unsure of extra term


    Rep:
    Hello again,

     \int \frac {1}{(s^2 - 0.01)^\frac{1}{2}} ds = ln (s^2 - 0.01+s)^\frac{1}{2} +C

    I can't find the rule for this one, not sure why there is an extra s appearing in there. Thanks so much for the help, it is appreciated!

    Rep:
    Hullo, I'm not sure what you mean, the 'extra' s is there because that's the integral, well aside the constant. Do you mean how do you do the integral? Well one way is a trig substitution as is usually the case when we have square roots and square floating about.

    Edit: I'm a fool, Ben-Smith is right, that's not the integral, nearly is though.

    Rep:
    (Original post by mh1985)
    Hello again,

     \int \frac {1}{s^2 - 0.01}^\frac{1}{2} \ds = ln (s^2 - 0.01+s)^\frac{1}{2}

    I can't find the rule for this one, not sure why there is an extra s appearing in there. Thanks so much for the help, it is appreciated!
    That's simply not right, just differentiate it to see.
    Also you haven't put the ds in or the +C.

    Rep:
    (Original post by mh1985)
    Hello again,

     \int \frac {1}{(s^2 - 0.01)^\frac{1}{2}} ds = ln (s^2 - 0.01+s)^\frac{1}{2} +C

    I can't find the rule for this one, not sure why there is an extra s appearing in there. Thanks so much for the help, it is appreciated!
    (Original post by ben-smith)
    That's simply not right, just differentiate it to see.
    Also you haven't put the ds in or the +C.

    OK sorry about that, edited the OP.

    Rep:
    (Original post by mh1985)
    OK sorry about that, edited the OP.
    Are you familiar with hyperbolic functions?

    Rep:
    (Original post by ben-smith)
    Are you familiar with hyperbolic functions?
    not quite 100% but I've heard of them

    Rep:
    You've got the 1/2 in the wrong place. It should be:

     \int \frac {1}{(s^2 - 0.01)^\frac{1}{2}} ds = ln ((s^2 - 0.01)^\frac{1}{2} +s) + C

    You can see that this is right by taking the derivative. The derivative of ln ((s^2 - 0.01)^\frac{1}{2}) + C = \frac{s}{s^2-0.01} (it's a chain rule party). The derivative of ln ((s^2 - 0.01)^\frac{1}{2} + s) + C, on the other hand, is \frac{1}{(s^2-0.01)^\frac{1}{2}}.

    Rep:
    (Original post by teamnoether)
    You've got the 1/2 in the wrong place. It should be:

     \int \frac {1}{(s^2 - 0.01)^\frac{1}{2}} ds = ln ((s^2 - 0.01)^\frac{1}{2} +s) + C

    You can see that this is right by taking the derivative. The derivative of ln ((s^2 - 0.01)^\frac{1}{2}) + C = \frac{s}{s^2-0.01} (it's a chain rule party). The derivative of ln ((s^2 - 0.01)^\frac{1}{2} + s) + C, on the other hand, is \frac{1}{(s^2-0.01)^\frac{1}{2}}.
    where does the other s come from? I doubt I would have got that without knowing the answer already. thanks for the reply btw.

    Rep:
    You need to make the substitution that s = 0.1sec(u). This probably seems pulled out of thin air , but comes from the fact that you have the square root of some number squared minus a constant, all inside a square root. Because sec^2(u) - 1 = tan^2(u), s^2- 0.01 would be 0.01sec^2(u) - 0.01 = 0.01(sec^2(u)-1) = 0.01tan^2(u).





(0.01tan^2(u))^\frac{1}{2} = 0.1tan(u).

    We also have ds = 0.1sec(u)tan(u), so our overall integral is

    \int sec(u)du because the tangents cancel.

    Is that enough to get you started? If not, let me know and I can keep going.

    Edit: Sorry - that came out slightly brusque upon rereading. It should have said something more like: If not, just let me know and I would be happy to keep going.

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: February 29, 2012

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