Integration that leads to a inverse trig identity

  • Thread starter Minutemade
  • Start date
  • #1
Ok, so a teacher showed an example in class awhile back. So im going over my notes right now, and I don't understand a certain part of the problem.

Also im new to the forums and its my first time posting here, so please push me in the right direction if i make a mistake.

integration of (x-2)(1/2)/(x+2)

the solution is: 2(x-2).5 - 4tan-1((x-2).5/2) + C

Basically what i tried doing, is:

u2 = x-2
u = (x-2).5
2u du = dx

which after a few steps leads me to:

2u - 8(integration of (u2 +4)-1)

After this i stop understanding the problem a little...
From here i sub the root x-2 back in as u and u2 as x-2, which gives me:

2(x-2).5 - 8ln|x+2| which is wrong i think....

The way the example continues is as such:
u2 = 4z2
u = 2z
du = 2 dz

2u - ((8)(2)/4) integration of dz/ (z2 +1)

which gives the answer given above...

Basically i dont understand why we want to sub in another letter for u2, and why i cant get the same answer when i sub in the (x-2).5 earlier.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Minutemade! Welcome to PF! :smile:
2u - 8(integration of (u2 +4)-1)

After this i stop understanding the problem a little...
From here i sub the root x-2 back in as u and u2 as x-2, which gives me:

2(x-2).5 - 8ln|x+2| which is wrong i think....

ah … you forgot to change the du as well. :wink:
Basically i dont understand why we want to sub in another letter for u2

You don't really need to …

most people wouldn't …

but it does stop you making a mistake if you can't remember the general formula! :smile:
 

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