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Trig integral help

  • Thread starter noblerare
  • Start date
47
0
1. Homework Statement

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

2. Homework Equations

n/a

3. The Attempt at a Solution

Here's what I've tried:
u=tanx
x=arctanu

dx=[tex]\frac{du}{1+u^2}[/tex]

[tex]\int[/tex][tex]\frac{du}{(1+u^2)(1-u^2)}[/tex]

I then used partial fractions to get:
[tex]\int[/tex][tex]\frac{0.5}{1-u^2}[/tex]+[tex]\frac{0.5}{1+u^2}[/tex]du

[tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{1+u^2}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int\frac{du}{1-u^2}[/tex]

The left side is simply arctanu while I used trig substitution to solve the right side
[tex]\frac{1}{2}[/tex]arctanu-[tex]\frac{1}{2}[/tex][tex]\int[/tex]csc[tex]\theta[/tex]d[tex]\theta[/tex]

[tex]\frac{1}{2}[/tex]arctan(tanx)+[tex]\frac{1}{2}[/tex]ln(csc[tex]\theta[/tex]+cot[tex]\theta[/tex]) + C

[tex]\frac{1}{2}[/tex]x + [tex]\frac{1}{2}[/tex]ln([tex]\frac{1}{\sqrt{1-u^2}}[/tex]+[tex]\frac{u}{\sqrt{1-u^2}}[/tex]) + C

With more re-substitution, I end up with:

[tex]\frac{1}{2}[/tex]x+[tex]\frac{1}{2}[/tex]ln(tanx+1)-[tex]\frac{1}{2}[/tex]ln([tex]\sqrt{1-tan(x)^2}[/tex]+C

For some reason, this is incorrect because when I try to take the derivative of that, I do not end up with [tex]\frac{1}{1-tan(x)^2}[/tex]

Is there something I'm doing wrong? What should I do?
 

Answers and Replies

1,750
1
I'd like to see your work with the partial fractions.

Did it look like ... [tex]\frac{1}{(1+u^2)(1-u^2)}=\frac{Ax+B}{1+u^2}+\frac{Cx+D}{1-u^2}[/tex] ... ?
 
rock.freak667
Homework Helper
6,230
31
I'd like to see your work with the partial fractions.

Did it look like ... [tex]\frac{1}{(1+u^2)(1-u^2)}=\frac{Au+B}{1+u^2}+\frac{Cu+D}{1-u^2}[/tex] ... ?
Shouldn't it be

[tex]\frac{1}{(1+u^2)(1-u)(1+u)} \equiv \frac{Au+B}{1+u^2}+\frac{C}{1-u}+ \frac{D}{1+u}[/tex]

or does it not matter about the [itex]1-u^2[/itex]?
 
1,750
1
Shouldn't it be

[tex]\frac{1}{(1+u^2)(1-u)(1+u)} \equiv \frac{Au+B}{1+u^2}+\frac{C}{1-u}+ \frac{D}{1+u}[/tex]

or does it not matter about the [itex]1-u^2[/itex]?
Good question, I overlooked that part.
 
exk
119
0
I think his decomposition is valid (it certainly combines back correctly).

Try using this: [tex]\int \frac{1}{a^{2}-u^{2}}du = \frac{1}{2a}\ln{\frac{|u+a|}{|u-a|}}+c[/tex]


For your sake I hope you are using a calculator when you work it out.
 
47
0
To rocomath: yes, my partial fractions work did look like that. I did not try it the way rock.freak667 did it. Does it matter?

To exk: Thanks for the help, but could you maybe explain how that you got that integral? And yes, I am using a calculator : D
 
Gib Z
Homework Helper
3,344
4
He got that integral through partial fractions again, and then using [itex]\log b - \log a = \log \left( \frac{b}{a}\right)[/itex]
 
benorin
Homework Helper
1,067
14
I think his decomposition is valid (it certainly combines back correctly).

Try using this: [tex]\int \frac{1}{a^{2}-u^{2}}du = \frac{1}{2a}\ln{\frac{|u+a|}{|u-a|}}+c[/tex]


For your sake I hope you are using a calculator when you work it out.
To do this integral, use partial fractions and integrate.
 
47
0
How do I make

[tex]\int[/tex][tex]\frac{dx}{1-tan^2x}[/tex] look like [tex]\int[/tex][tex]\frac{du}{a^2-u^2}[/tex] though?

Do I need to subtitute u=tanx? If so, I don't know how to change the variable of integration so that it matches that the second integral

Or can I simply take tanx as plug it into that equation?
 
benorin
Homework Helper
1,067
14
Wow! noblerare, your solution is correct!

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

Check:

[tex]\frac{d}{dx}\left[\frac{1}{2}x+\frac{1}{2}\ln\left( 1+ \tan x\right)-\frac{1}{2}\ln \sqrt{1- \tan^2 x}+C\right] = \frac{1}{2}+\frac{\sec^2 x}{2(1+ \tan x)}+\frac{\sec^2 x\tan x}{2(1- \tan^2 x)} = \frac{1}{1- \tan^2 x}[/tex]
 
exk
119
0
How do I make

[tex]\int[/tex][tex]\frac{dx}{1-tan^2x}[/tex] look like [tex]\int[/tex][tex]\frac{du}{a^2-u^2}[/tex] though?

Do I need to subtitute u=tanx? If so, I don't know how to change the variable of integration so that it matches that the second integral

Or can I simply take tanx as plug it into that equation?
You already had it in that form following your substitution. You have:

[tex]
\frac{1}{2}\int\frac{du}{1+u^2}+\frac{1}{2}\int\frac{du}{1-u^2}
[/tex]

If a=1 and u=u, then the integral in the second part is in the form you want.
 
47
0
ah...i see,i got it; okay, thanks everybody!
 

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