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integral question - unsure of extra term

Hello again,

1(s20.01)12ds=ln(s20.01+s)12+C \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!
(edited 12 years ago)
Reply 1
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.
(edited 12 years ago)
Original post by mh1985
Hello again,

Unparseable latex formula:

\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.
Reply 3
Original post by mh1985
Hello again,

1(s20.01)12ds=ln(s20.01+s)12+C \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.
(edited 12 years ago)
Original post by mh1985
OK sorry about that, edited the OP.


Are you familiar with hyperbolic functions?
Reply 5
Original post by ben-smith
Are you familiar with hyperbolic functions?


not quite 100% but I've heard of them
Reply 6
You've got the 1/2 in the wrong place. It should be:

1(s20.01)12ds=ln((s20.01)12+s)+C \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((s20.01)12)+C=ss20.01ln ((s^2 - 0.01)^\frac{1}{2}) + C = \frac{s}{s^2-0.01} (it's a chain rule party). The derivative of ln((s20.01)12+s)+Cln ((s^2 - 0.01)^\frac{1}{2} + s) + C, on the other hand, is 1(s20.01)12\frac{1}{(s^2-0.01)^\frac{1}{2}}.
Reply 7
Original post by teamnoether
You've got the 1/2 in the wrong place. It should be:

1(s20.01)12ds=ln((s20.01)12+s)+C \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((s20.01)12)+C=ss20.01ln ((s^2 - 0.01)^\frac{1}{2}) + C = \frac{s}{s^2-0.01} (it's a chain rule party). The derivative of ln((s20.01)12+s)+Cln ((s^2 - 0.01)^\frac{1}{2} + s) + C, on the other hand, is 1(s20.01)12\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.
Reply 8
You need to make the substitution that s = 0.1sec(u). This probably seems pulled out of thin air :smile:, but comes from the fact that you have the square root of some number squared minus a constant, all inside a square root. Because sec2(u)1=tan2(u),s20.01sec^2(u) - 1 = tan^2(u), s^2- 0.01 would be 0.01sec2(u)0.01=0.01(sec2(u)1)=0.01tan2(u).[br][br][br](0.01tan2(u))12=0.1tan(u)0.01sec^2(u) - 0.01 = 0.01(sec^2(u)-1) = 0.01tan^2(u).[br][br][br](0.01tan^2(u))^\frac{1}{2} = 0.1tan(u).

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

sec(u)du\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.
(edited 12 years ago)

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