Question

Evaluate the integral of 9xarctanx dx from 1 to 0

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Then it is not from 1 to 0

Answer 5

yeah, sorry

Answer 6

all for you @Psymon

Answer 7

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?

Answer 8

yes, I am

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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 : )

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

okay, that's a good suggestion

Answer 19

as you said, it just creates more numbers

Answer 20

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 : )

Answer 21

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

Answer 22

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

Answer 23

are you just trying to simplify it or what?

Answer 24

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

Answer 25

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

Answer 26

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).

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Thank you! HUGE help!

Answer 34

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