Okay, I have a question. Is equal to ?

I am working on this problem,

, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.

I made

Am I heading in the right direction? Thanks in advance.

-Matt

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- Sep 3rd 2008, 12:10 AMmatt3DSimple Integral question w/ parts
Okay, I have a question. Is equal to ?

I am working on this problem,

, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.

I made

Am I heading in the right direction? Thanks in advance.

-Matt - Sep 3rd 2008, 12:33 AMflyingsquirrel
Hello,

No, we have or .

Quote:

I am working on this problem,

, it's actually tan^-1(2x) but I'm not sure how to insert the negative to make it the inverse.

Quote:

I made

Am I heading in the right direction?

Does it help ? - Sep 3rd 2008, 12:50 AMProve It
- Sep 3rd 2008, 08:39 PMmatt3D
Oops, I mixed my problem up. This is what I am currently trying to figure out:

- Sep 3rd 2008, 09:00 PMmatt3D
Okay so, correct? This is looking more difficult then I thought! So now: I put the in front of the integral to get rid of the 2 from du which is 2dy...

- Sep 3rd 2008, 09:22 PMChris L T521
The terms in red should be 2, not 4.

Now, from here, do the following:

We found that

The integral reappears on the right side! So get it on the left side!

Before we get our answer, try to evaluate the integral I highlighted in red.

Hint : apply integration by parts! Let ..... - Sep 3rd 2008, 09:32 PMo_O

First, let's make the sub:

and we get:

Now you can use integration by parts: - Sep 3rd 2008, 09:40 PMmatt3D
Yea, I see what you mean! (Nod)

How does it become 3/2 on the left side? I know its because you put the integral to the left though like you said, but how exactly?

Okay, I never thought of it like that, I'm going to try that now. - Sep 3rd 2008, 09:46 PMChris L T521
- Sep 3rd 2008, 09:50 PMmatt3D
- Sep 3rd 2008, 09:58 PMmatt3D
- Sep 3rd 2008, 10:03 PMChris L T521
- Sep 3rd 2008, 10:07 PM11rdc11
Try

oops I was too slow - Sep 3rd 2008, 10:24 PMmatt3D
Alright, so I finally get: Now I'm a little confused about the division though, 11rdc11, could you explain your last post a little more?

Thank you guys for all of your help! I really do appreciate it. - Sep 3rd 2008, 10:28 PMChris L T521