Evaluate the integral of 9xarctanx dx from 1 to 0

- anonymous

Evaluate the integral of 9xarctanx dx from 1 to 0

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- schrodinger

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- anonymous

this one?\[I=9\int\limits_{1}^{0}x·\arctan(x)dx\]

- anonymous

\[\int\limits_{0}^{1} 9x \tan^{-1} x \]

- Psymon

We usually say the bottom limit first. So this would be 0 to 1. So thats why he wrote it backwards :P

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- anonymous

Then it is not from 1 to 0

- anonymous

yeah, sorry

- anonymous

all for you @Psymon

- Psymon

This would be integration by parts. You would let x be your u (we can multiply in the 9 later) and arctan(x) be your v. So that means we follow this formula:
\[u \int\limits_{}^{}v- [\int\limits_{}^{}u'\int\limits_{}^{}v]\]So if we allow u to be x, then u' is simply 1. If we allow v to be arctan(x),then \[\int\limits_{}^{}v=\frac{ 1 }{ x ^{2}+1} \]
Kinda with me so far?

- anonymous

yes, I am

- Psymon

I apologize, I chose the wrong u and v. The process is the same, I just needed arctan(x) to be the u and x to be the v. I'll start over :P
So if u is arctan(x), then du is:
\[\frac{ 1 }{ x ^{2}+1 } \]and if v = x, then \[\int\limits_{}^{}v = \frac{ x ^{2} }{ 2 } \]
I apologize for that.

- Psymon

So let's plug in those values:
\[\frac{ x ^{2} }{ 2 }arctanx - [\int\limits_{}^{}\frac{ x ^{2} }{ 2(x ^{2}+1) }dx] \]
This would be the proper form after plugging in values based on the formula

- anonymous

wouldn't v=9/2x^2?

- Psymon

Well, Im saving the 9 for later. I can multiply it in at the end. Asof now, its just another number that may get in the way. I am not sure how you decided on your v, though. Can you explain? That way we can help get on the same page : )

- Psymon

Oh, did you accidentally put the x^2 in the denominator of your suggestion or did you intend for it to be therE?

- anonymous

\[9/2x ^{2}\tan^{-1} -9/2\int\limits_{0}^{1}x ^{2}/1+x ^{2}\]

- Psymon

You make it look like you intend to have x^2 in the denominator. It should be in the numerator. If it's in the numerator then what you have is correct.

- anonymous

yeah, I didn't mean for it to be in the denominator

- Psymon

Ah, okay. Then yes, you are fine. I just decided to save the 9 for later : )

- anonymous

okay, that's a good suggestion

- anonymous

as you said, it just creates more numbers

- Psymon

So yeah, I will factor out the 1/2 like you did on yours.
\[\frac{ x ^{2} }{ 2 }\arctan(x) - \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ x ^{2} }{ x ^{2}+1 }dx \]
Now what we have remaining can be done by long division, but I prefer a faster way. What I will do is add 1 and subtract one from the numerator at the same time. After I do this, I can split our fraction into 2:
\[\frac{ x ^{2} }{ 2 }\arctan(x) - \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ x ^{2}+1-1 }{ x ^{2}+1 }dx \]
\[\frac{ x ^{2} }{ 2 }\arctan(x) - \frac{ 1 }{ 2 }[\int\limits_{}^{}1dx-\int\limits_{}^{}\frac{ 1 }{ x ^{2}+1 }dx]\]
This allows us to integrate like normal now. Of course, the integration of 1 is just x. And the integration of our 2nd integral bring us back to arctan(x) This means I now have this:
\[\frac{ x ^{2} }{ 2 }\arctan(x) - \frac{ 1 }{ 2 }(x - arctanx)\]From here we can finally plug in our limits : )

- Psymon

Well, not forgetting the 9 we factored out xD
\[\frac{ 9x ^{2} }{ 2 }\arctan(x) - \frac{ 9 }{ 2 }(x-arctanx)\]

- anonymous

so it's the first part when you split the fraction that I'm not understanding

- anonymous

are you just trying to simplify it or what?

- Psymon

Yes. There are two methods of handling the splitting. Normally we cannot integrate it as is, but as a normal recognition, when you see that the numerator and denominator are of the same degree, you always have the option of long division. SO it's something to maybe have in the back of your mind. But what I did was something a little bit faster. It allowed me to get the same result, but using a little algebra trick. Adding and subtracting by 1 is essentially addition by 0. But by adding and subtracting the one, I am ableto split it up into these two fractions:
\[\frac{ x ^{2}+1-1 }{ x ^{2}+1 } \]becomes
\[\frac{ x ^{2}+1 }{ x ^{2}+1 }-\frac{ 1 }{ x ^{2}+1 } \]
Any term in the numerator I can choose to make a fraction out of it. So if I wanted to, I could make that into 3 fractions because of 3 terms in the numerator. But my little trick allowed me to rewrite it like so. If you do long division instead, you'll get the same answer.
1
x^2+1|x^2
-(x^2 + 1)
-1 remainder

- anonymous

Nice! So you made the first fraction equal "1". Makes sense now.

- Psymon

Yep, absolutely :3 It can be done with a bunch of things, it's pretty neat actually. But yes, that allows us to get to the answer we have above. Now its just doing our regular plugging in limits, doing F(1) - F(0).

- Psymon

\[\frac{ 9(1)^{2} }{ 2 }\arctan(1) - \frac{ 9 }{ 2 }(1 - \arctan(1))\]

- Psymon

arctan(1) is pi/4 if we remember much about the unit circle from way back in trig. So that means we have:
\[\frac{ 9\pi }{ 8 }-\frac{ 9 }{ 2 }(1-\frac{ \pi }{ 4 })\]So now we subtract F(0) from this:
\[-(\frac{ 9(0)^{2} }{ 2 }\arctan(0) - \frac{ 9 }{ 2 }(0-\arctan(0))\]arctan(0) is where sin is 0 and cosine is defined, meaning 0. So essentially, F(0) is all 0, meaning the answer is just what I have above.

- Psymon

\[\frac{ 9\pi }{ 8 }-\frac{ 9 }{ 2 }+\frac{ 9\pi }{ 8} \]
\[\frac{ 18\pi }{ 8 }-\frac{ 9 }{ 2 }\]
\[\frac{ 9\pi }{ 4 }-\frac{ 9 }{ 2 } \]
Not a badidea to check my math, but this is the process :3

- Psymon

I have to head out now unfortunately. Hopefully all of this helped ^_^

- anonymous

oh, okay. I was also forgetting to distribute the negative so the radian was cancelling.

- Psymon

Yeah, which would be a lot of work for such a simple answer xD But yep, so that should be your answer. Good luck ^_^

- anonymous

Thank you! HUGE help!

- anonymous

I have compiled the whole process @psymon has gone through. Find attached

##### 1 Attachment

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