Everybody Loves P5 Integration! Watch

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Yttrium
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#1
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Everyone except me, that is...

If anyone could help me out with working to the following questions (I have answers but I don't know where they've come from)...

1) PROVE INT xarcoshx dx = 1/4(2x^2 -1)arcoshx - x/4(x^2 -1)^2 + C

I know you have to do this by parts, but I end up with:

x^2/2arcoshx - x - arcoshx ( I must be doing something wrong... )

2) INT (tanh5x)^2 dx (answer = x-tanh5x/5)

I just have no idea even how to start this one (but I do have a sneaky suspicion it might involve half-angle formulae in it somewhere - wish I understood them!)

3) INT (e^x +1)^-1 dx (answer = ln [e^x/(e^x +1)] )

I don't know how to start this one either; you can't do bracket-to-a-power integration as it's not linear inside the bracket... perhaps something like + 1 then -1; but I can't see anything that'd work. I'm probably missing something obvious.

4) INT (2x+1)/(x^2 - 9)

Now, I reckon partial fractions are involved in this one, and I got them as follows:

INT 7/6(x-3) - 5/6(x+3) dx (but this might be wrong)

This gives me something like 7/6ln|x-3| - 5/6ln|x+3|
but the answer is supposed to be ln|x^2 - 9| + 1/6ln |x-3|/|x+3|

Even if I rearrange my logs, the first term will always be missing. Hmmm...
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m:)ckel
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4) INT (2x+1)/(x2-9) dx

= INT [(2x)/(x2-9)] dx + INT [1/(x2-9)] dx

= ln(x2-9) + (1/6)ln[(x-3)/(x+3)] + c

(the first, because derivative of the bottom is on top, and second is a standard result)
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Gaz031
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#3
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In number 2 you use the identity 1-(tanh5x)^2=(sech5x)^2, as you can integrate 1 and (sech5x)^2.

Q4: http://img105.echo.cx/img105/7057/swscan000158cf.jpg
Q3: http://img105.echo.cx/img105/7422/swscan000167gx.jpg
Q2: Use the identity 1-(tanh5x)^2=(sech5x)^2, as you can integrate 1 and (sech5x)^2.
Q1: http://img105.echo.cx/img105/1315/swscan000170zx.jpg
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m:)ckel
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3) Try the substitution u=ex+1
You'll get INT 1/[(u)(u-1)] du , which you can do partials on. Not sure if it's the quickest way, but a start nonetheless.
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Spenceman_
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For number 3) first write it as a fraction and then divide top and bottom by ex . You'll recognise the fraction your left with as a "ln" hopefully. And then simplify the result to get the form the answer's given in.

Spence
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Yttrium
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#6
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Great!

Thanks everyone (especially Gaz)!

I expect I might reuse this thread if (more likely when) I come across even yuckier integrals in the future.

Thanks!

--

Yttrium
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Datura
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#7
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I'm having a dumb moment, can somone remind me how to integrate (sech5x)^2?
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Jonny W
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#8
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(Original post by Bebop)
I'm having a dumb moment, can somone remind me how to integrate (sech5x)^2?
(1/5)tanh(5x) + c

Rep for anyone who can integrate sech(x).
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Gaz031
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(Original post by Bebop)
I'm having a dumb moment, can somone remind me how to integrate (sech5x)^2?
You know that the derivative of tanhx is (sechx)^2.
INT (sech5x)^2 dx, u=5x, du/dx=5, dx=(1/5)du.
INT (sech5x)^2 dx = (1/5)INT (sechu)^2 du = (1/5)(tanhu)+C=(1/5)(tanh5x)+C
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Gaz031
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(Original post by Jonny W)
(1/5)tanh(5x) + c

Rep for anyone who can integrate sech(x).
INT sechx dx = INT (1/coshx) dx = 2INT1/(e^x+e^-x)=2INT[e^x/(1+e^2x)] dx
u=e^x transforms the integral into: 2INT1/(1+u^2) du = 2arctanu+C=2arctan(e^x)+C.
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Spenceman_
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#11
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In the same way tanx differentiate's to sec2 x

sech2 x will integrate to tanhx

sech2 5x will integrate to (1/5)tanhx
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Spenceman_
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#12
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hmm took too long with that post lol.
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dvs
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#13
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Another way to do it:

I = ∫ sech(x) dx

t = tanh(x/2)
dt/dx = (1/2)sech²(x/2) = (1-t²)/2
dx/dt = 2/(1-t²)
cosh(x) = cosh²(x/2) + sinh²(x/2) = (1+t²)/(1-t²)
sech(x) = (1-t²)/(1+t²)

I = ∫ 2/(1+t²) dt
= 2 arctan(t) + C
= 2 arctan(tanh(x/2)) + C
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Datura
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#14
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(Original post by dvs)
Another way to do it:

I = ∫ sech(x) dx

t = tanh(x/2)
dt/dx = (1/2)sech²(x/2) = (1-t²)/2
dx/dt = 2/(1-t²)
cosh(x) = cosh²(x/2) + sinh²(x/2) = (1+t²)/(1-t²)
sech(x) = (1-t²)/(1+t²)

I = ∫ 2/(1+t²) dt
= 2 arctan(t) + C
= 2 arctan(tanh(x/2)) + C
I like this.
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Aitch
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#15
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(Original post by Yttrium)
Everyone except me, that is...

If anyone could help me out with working to the following questions (I have answers but I don't know where they've come from)...

1) PROVE INT xarcoshx dx = 1/4(2x^2 -1)arcoshx - x/4(x^2 -1)^2 + C

I know you have to do this by parts, but I end up with:

x^2/2arcoshx - x - arcoshx ( I must be doing something wrong... )

2) INT (tanh5x)^2 dx (answer = x-tanh5x/5)

I just have no idea even how to start this one (but I do have a sneaky suspicion it might involve half-angle formulae in it somewhere - wish I understood them!)

3) INT (e^x +1)^-1 dx (answer = ln [e^x/(e^x +1)] )

I don't know how to start this one either; you can't do bracket-to-a-power integration as it's not linear inside the bracket... perhaps something like + 1 then -1; but I can't see anything that'd work. I'm probably missing something obvious.

4) INT (2x+1)/(x^2 - 9)

Now, I reckon partial fractions are involved in this one, and I got them as follows:

INT 7/6(x-3) - 5/6(x+3) dx (but this might be wrong)

This gives me something like 7/6ln|x-3| - 5/6ln|x+3|
but the answer is supposed to be ln|x^2 - 9| + 1/6ln |x-3|/|x+3|

Even if I rearrange my logs, the first term will always be missing. Hmmm...
Q1.

After integration by parts, when you get to needing

∫x² /√(x²-1) dx

you can -1+1 to the numerator to get

∫(x²-1+1) /√(x²-1) dx

split to get

∫(x²-1) /√(x²-1) dx +∫1 /√(x²-1) dx

∫√(x²-1) dx +∫1 /√(x²-1) dx

Then sub for x (in the first integral).

This is probably done by someone else in here somewhere, but never mind!

Aitch
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