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integration problem

dibs101

can anyone give me a dig out with this problem...

find indefinate integral of arctan x/1 + x^2

I realise that the integral of 1/1 +x^2 is arctan x... but where next?

Anyhelp greatly appreciated...

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Iwasfrozen

[Integrate](x^2 + tan^-1 (x)) =
[Integrate](x^2) + [Integrate](tan^-1(x) =
[1/3.(x^3)] + [Integrate](tan^-1(x)]
Then we integrate tan^-1(x) on it's own.
u = tan^-1(x)
du = 1/(x^2 + 1).dx
dv = 1
v = x.dx
Then we use the u.v - [integrate][v.du] =
tan^-1(x).x - [integrate][x/(x^2 + 1)]
To integrate [x/(x^2 + 1)]
let u = x^2 + 1
du = 2x =
1/2[integrate] 1/u.du =
1/2ln[x^2 + 1]
Put that back into the origional equation and you get:
[integrate]tan^-1 (x) = tan^-1(x).x -1/2ln[x^2 + 1]
Put that back into the origional quation and you get:
1/3[x^3] + tan^-1(x).x -1/2ln[x^2 + 1] + c
That was a hard question, are you in college or doing LC?
Mods:I know I'm not supposed to do it out compleatly but it was a hard question.

[Deleted User]

You want to find [latex]\displaystyle\int\arctan\left(\frac{x}{1 + x^2}\right)[/latex], yes?

Use integration by parts, letting u equal [latex]\displaystyle\arctan\left(\frac{x}{1 + x^2}\right)[/latex], and letting v equal 1.

You'll have to use a substitution to differentiate [latex]\displaystyle\frac{x}{1 + x^2}[/latex]. Let w equal [latex]\mbox{1} + x^2}[/latex], and find dw etc.

Use the integration by parts formula:

[latex]\int\mbox{udv} = \mbox{uv} - \int\mbox{vdu}[/latex]

If you've any specific problems just say.

Iwasfrozen

:eek:
I need to figure out how to use latex.

[Deleted User]

Iwasfrozen wrote: »

:eek:
I need to figure out how to use latex.

It only takes a few minutes to get used to it. Have this open beside you when you try to use it:

http://www.physicsforums.com/misc/howtolatex.pdf

And

http://amath.colorado.edu/documentation/LaTeX/Symbols.pdf

On Boards you just wrap the LaTeX code with [.latex] [./latex] (without the full stops).

Iwasfrozen

[latex] \int \sqrt{16-(\4sinx)^2} = \int \sqrt{(4)^2-(\4sinx)^2} [/latex]
[latex] 16 - 16\sin^2x = 16(1 - \sin^2x) = 4\cos^2x[/latex]
[latex] \int \sqrt{4cos^2x} = \int 2cosx [/latex]
[latex] \int 2cosx = 2sinx + c [/latex]

[Deleted User]

Also, just a tip, the LaTeX looks "nicer" if you use \displaystyle.

For example:

[latex]\int\sqrt{16 - \sin^2(x)}\ \mbox{dx}[/latex]

Becomes:

[latex]\displaystyle\int\sqrt{16 - \sin^2(x)}\ \mbox{dx} = \mbox{4E}\left(x\vert\frac{1}{16}\right) + C[/latex]

P.S. The integral you've carried out is also an elliptical integral.

Iwasfrozen

Where do you put the \displaystyle ?
Also whats an elliptical integral ?

[Deleted User]

Iwasfrozen wrote: »

Where do you put the \displaystyle ?

You put it in at the beginning of the LaTeX code, for example: [.latex]\displaystyle\sqrt{\sin(x)}[./latex], again without the full stops.

Also whats an elliptical integral?

I'm sure one of the college-level contributors around here would be able to answer this question an awful lot better than I can, as I know very little about them.

From what I understand, any elliptical integral can be defined as an integral with respect to x of a rational funtion of x, g(x), and sqrt(f(x)), where f(x) is a polynomial of the 3rd of 4th (or even more?) degree. Integrals of this kind can be reduced by a change of variables (i.e. a substitution) to a sum of basic functions and/or elliptic integrals of the first, second, and/or third kinds. (There are distinct differences between the three kinds, but it's rather technical and I'm not 100% sure myself of the differences).

The problem above was an elliptical integral of the second kind; and basically an elliptical integral is any integral which takes a certain form, such as this (which is a second kind variation):

[latex]\displaystyle\mbox{E(\varphi\mid\mbox{m}) = \int_{0} ^\varphi \sqrt{1 - m\sin^2(t)} \ \mbox{dt}[/latex]

Edit
I'm just going to clarify a little bit more.

In your example, [latex]\displaystyle\int\sqrt{16 - \sin^2(x)}\ \mbox{dx}[/latex], you can factor out the 16, which leaves you with:

[latex]\displaystyle\int\sqrt{16\left(1 - \frac{1}{16}\sin^2(x)\right)}\ \mbox{dx}[/latex]

You can then remove the 16, which turns into a 4 outside of the squareroot, leaving you with:

[latex]\displaystyle\mbox{4}\int\sqrt{\left(1 - \frac{1}{16}\sin^2(x)\right)}\ \mbox{dx}[/latex]

Now, if you compare this to the example I've given, you'll notice that [latex]\displaystyle\mbox{m} = \frac{1}{16}[/latex], leaving you with the solution - by comparison - of [latex]\displaystyle\mbox{4E}\left(x|\frac{1}{16}\right)[/latex].

Finding the full solution is the trickest part, and I've left that out as it's a bit complicated.

End of edit.

From everything I've read on them, the notation is the most confusing (for me personally). I find it hard to get my head around. As for what their uses are? I'm not exactly sure.

If anyone notices any mistakes in my definition please feel free to correct what I've said, as all I've learned about elliptical integrals has came from a downloadable .pdf document, which as you've probably guessed isn't very much. If anybody has anything to add I'd love to know more about them.

Iwasfrozen

I'm still not exactly sure what you're saying.
Do you mean that by substituting [latex]\displaystyle x = 4sinx [/latex]
to get [latex] \displaystyle \sqrt {16(1-sin^2x)} = \sprt{16cos^2} [/latex]
Oh wait, i see my mistake it should have been [latex]\displaystyle{16cos^2x}[/latex] rather than [latex]\displaystyle{4cos^2}[/latex]

Time Magazine

If you want to avoid typing the full stop [.latex] then you can use the [noparse][noparse][/noparse][/noparse] tags to, well, stop Boards from parsing the BB code.

ray giraffe

Hmm, I think the OP meant (arctan x)/(1+x^2), which would be much easier to integrate - just use integration by substitution with u=arctan x