Solving Trig Integral: \int\frac{dx}{1-tan^2(x)}

[noblerare]

Homework Statement

\int\frac{dx}{1-tan^2(x)}

Homework Equations

n/a

The Attempt at a Solution

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

dx=\frac{du}{1+u^2}

\int\frac{du}{(1+u^2)(1-u^2)}

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

\frac{1}{2}\int\frac{du}{1+u^2}+\frac{1}{2}\int\frac{du}{1-u^2}

The left side is simply arctanu while I used trig substitution to solve the right side
\frac{1}{2}arctanu-\frac{1}{2}\intcsc\thetad\theta

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

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

With more re-substitution, I end up with:

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

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

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

 

[rocomath]

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

Did it look like ... \frac{1}{(1+u^2)(1-u^2)}=\frac{Ax+B}{1+u^2}+\frac{Cx+D}{1-u^2} ... ?

 

[rock.freak667]

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

Did it look like ... \frac{1}{(1+u^2)(1-u^2)}=\frac{Au+B}{1+u^2}+\frac{Cu+D}{1-u^2} ... ?

Shouldn't it be

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

or does it not matter about the 1-u^2?

 

[rocomath]

Shouldn't it be

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

or does it not matter about the 1-u^2?

Good question, I overlooked that part.

 

[exk]

I think his decomposition is valid (it certainly combines back correctly).

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


For your sake I hope you are using a calculator when you work it out.

 

[noblerare]

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]

He got that integral through partial fractions again, and then using \log b - \log a = \log \left( \frac{b}{a}\right)

 

[benorin]

I think his decomposition is valid (it certainly combines back correctly).

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


For your sake I hope you are using a calculator when you work it out.

To do this integral, use partial fractions and integrate.

 

[noblerare]

How do I make

\int\frac{dx}{1-tan^2x} look like \int\frac{du}{a^2-u^2} 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]

Wow! noblerare, your solution is correct!

\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

Check:

\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}

 

[exk]

How do I make

\int\frac{dx}{1-tan^2x} look like \int\frac{du}{a^2-u^2} 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:

<br /> \frac{1}{2}\int\frac{du}{1+u^2}+\frac{1}{2}\int\frac{du}{1-u^2}<br />

If a=1 and u=u, then the integral in the second part is in the form you want.

 

[noblerare]

ah...i see,i got it; okay, thanks everybody!