Integral with square root

  • Thread starter twoflower
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  • #1
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Hi,

I'm trying to find this integral:

[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx
[/tex]

Because [itex]1-x^2[/itex] has two different real solutions, I can write

[tex]
\sqrt{ax^2 + bx + c} = \sqrt{-a}(x_2 - x)\sqrt{\frac{x-x_1}{x_2 - x}}
[/tex]
so
[tex]
\sqrt{1-x^2} = (-1 - x)\sqrt{\frac{x - 1}{-1 - x}} = (-1 - x)\sqrt{\frac{1-x}{1+x}}
[/tex]

I used this substitution:
[tex]
t = \sqrt{\frac{1-x}{1+x}}
[/tex]

It gives
[tex]
x = \frac{1-t^2}{1 + t^2}
[/tex]

[tex]
x + 1 = \frac{2}{1+t^2}
[/tex]

[tex]
dx = \frac{-4t}{(1+t^2)^2}
[/tex]

So
[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \int \frac{ \frac{1-t^2}{1+t^2}}{\left(\frac{2}{1+t^2}\right)\left(\frac{-2}{1+t^2}\right)t}\ \ \frac{-4t}{(1+t^2)^2}\ dt
[/tex]
[tex]
= \int \frac{1-t^2}{1+t^2}\ dt = \int \frac{1}{1+t^2}\ dt - \int \frac{t^2}{1+t^2}\ dt = \arctan t -\ \int \frac{t^2}{1+t^2}\ dt
[/tex]

Damn I know I should be able to solve this integral, but I don't know how, maybe it's too late for me... :frown:

Btw the correct result should be:

[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \sqrt{\frac{1-x}{1+x}} + 2\arctan \sqrt{\frac{1+x}{1-x}} + C
[/tex]

Thank you.
 
Last edited:

Answers and Replies

  • #2
GCT
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Why don't you try a simple trig substitution and go from there? You can first get rid of the square root and try to simplify.

[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx
[/tex]

[tex] \int \frac{sin@cos@~d@}{(sin@cos@+cos@)} [/tex]

Next try multiplying by [tex] \frac{(cos@-sin@cos@)}{cos@-sin@cos@} [/tex]
 
  • #3
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Thank you GCT, it's interesting idea, but it seems it leads to [itex]t = \tan \frac{x}{2}[/itex] substitution, which is little tricky in my experience...could you look please at my original solution and tell me what's wrong there?

Thanks!
 
  • #4
dextercioby
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A quick glance

[tex] x+1=\frac{2}{1+t^{2}} [/tex]

,assuming "x" was right in the first place.

Daniel.
 
  • #5
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dextercioby said:
A quick glance

[tex] x+1=\frac{2}{1+t^{2}} [/tex]

,assuming "x" was right in the first place.

Daniel.
Thank you Daniel! How stupid I am..So I'm gonna go solve it again, hopefully it will be ok now :smile:
 
  • #6
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I edited my initial post now, could you now help me how to solve that?
 
  • #7
dextercioby
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[tex] \int \frac{t^{2}}{1+t^{2}} \ dt=\int \frac{1+t^{2}-1}{1+t^{2}} \ dt [/tex]

Daniel.
 
  • #8
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Thank you I noticed it already :) Now the result seems ok, except for sign and similar technicalities..
 
  • #9
dextercioby
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You can check the sign out.It looks okay to me.


Daniel.
 
  • #10
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Well, the right result should be
[tex]
\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \sqrt{\frac{1-x}{1+x}} + 2\arctan \sqrt{\frac{1+x}{1-x}} + C
[/tex]

but I have
[tex]
2 \arctan \sqrt{\frac{1-x}{1+x}} - \sqrt{\frac{1-x}{1+x}}
[/tex]
 
  • #11
dextercioby
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I don't have time now,i'll take a deeper look later.

Daniel.
 
  • #12
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[tex]\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx=\int \frac{x+1-1}{(x+1)\sqrt{1-x^2}}\ dx=\int \frac{1}{\sqrt{1-x^2}}\ dx-\int \frac{1}{(x+1)\sqrt{1-x^2}}\ dx[/tex]
First one is obvious. For second [tex]t=\frac{1}{x+1}[/tex] will lead to simple solution.
The answer then is different from what you've given, but Mathematica says that it's absolutely correct.
 
  • #13
GCT
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Twoflower, in case you're interested

[tex] \int \frac{sin@cos@~d@}{(sin@cos@+cos@)} * \frac{(cos@-sin@cos@)}{cos@-sin@cos@} [/tex]

[tex] \int \frac{sin@cos^{2}@-sin^{2}cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@} [/tex]

[tex] \int \frac{sin@cos^{2}@}{cos^{2}@-sin^{2}@cos^{2}@}~-~ \int \frac{sin^{2}@cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@}[/tex]

[tex] \int \frac{sin@~d@}{cos^{2}@}~-~ \int \frac{sin^{2}@~d@}{cos^{2}@} [/tex]

The left term can be resolved using substitution. The right term simplifies to
[tex] \int \frac{1}{cos^{2}@}~-~\int d@ [/tex]

anyone see any mistakes? Please point them out.
 
  • #14
147
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[tex]\int \frac{sin@~d@}{cos^{2}@} = -\int \frac{d(\cos@)}{cos^{2}@}[/tex]

It's all right with your solution GCT. At least it's not so artificiall as which Twoflower made.
 

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