Trigonometric Integral Confusion

  • #1
Char. Limit
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Homework Statement



So, for absolutely no good reason I decided to practice trigonometric substitution. I started with [itex]\int \left(1-x^2\right)^{-1/2} dx[/itex], and that was easy, everything cancelled out nicely. Then I tried [itex]\int \left(1+x^2\right)^{-1/2} dx[/itex], and although the integral was fine, everything did not cancel out nicely afterward. So I need help on one minor issue.

Homework Equations



[tex] sec(u) = \frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)}[/tex]

The Attempt at a Solution



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

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

[tex]\int sec(u) du[/tex]

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

[tex]\int \frac{dv}{v}[/tex]

[tex]ln(v)[/tex]

[tex]ln(tan(u) + sec(u))[/tex]

[tex]ln(x + sec(tan^{-1}(x)))[/tex]



Now, is there any way to simplify sec(arctan(x))?
 

Answers and Replies

  • #2
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Let θ = arctan(x), then tan(θ) = x = x/1. Draw a right triangle, label the sides, and then find sec(θ).
 
  • #3
vela
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Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1.
 
  • #4
Char. Limit
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Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1.

Hmm... I did that, but I don't think my answer is right... the O=x, A=1, H=sqrt(1+x^2) idea gives me sec(arctan(x))=sqrt(1+x^2), and that just doesn't sound right... but it probably is.

So...

[tex]ln\left(x+\sqrt{1+x^2}\right) +C[/tex]

Is the answer? For some reason I expected a trig function to pop up like it did with 1/sqrt(1-x^2)...
 
  • #5
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That's the right answer, you can always check on Wolfram|Alpha too.
 
  • #6
vela
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Or you could differentiate your answer and see if you get the integrand back.

If you think in terms of the hyperbolic trig functions, your answer is arcsinh(x), so a trig function does actually pop up.
 
  • #7
Char. Limit
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Oh I see. Thanks for the info on the arsinh(x) function... I just looked it up on Wikipedia, and it helped me understand that there was a trig function hidden in that logarithmic function.
 
  • #8
hunt_mat
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There are two ways to solve this integral. There is the use of [tex]x=\tan\theta[/tex] and by the use of the identity [tex]1+\tan^{2}theta =\sec^{2}\theta[/tex] and the use of [tex]x=\sinh u[/tex] and use [tex]\cosh^{2}u=1+\sinh^{2}u[/tex]. I personally think it is easier with the hyperbolic function.
 
  • #9
Char. Limit
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Yes, but I had never heard of the hyperbolic functions before today, so I don't quite know the derivative of sinh and cosh. Is it the same as the derivative of sin and cos? Namely...

[tex]\frac{d sinh(x)}{dx} = cosh(x)[/tex]

[tex]\frac{d cosh(x)}{dx} = - sinh(x)[/tex]

Are those correct?
 
  • #10
hunt_mat
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No. The definitions are as follows:
[tex]
\sinh x=\frac{e^{x}-e^{-x}}{2}\quad\cosh x=\frac{e^{x}+e^{-x}}{2}
[/tex]
From these definitions, you can calculate the derivatives and also prove the basic identity:
[tex]
\cosh^{2}x-\sinh^{2}x=1
[/tex]
From these three definitions, you can pretty much prove everything you need.

Mat
 
  • #11
Char. Limit
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Oh, I see. So the derivative is a two-point cycle, not a four-point cycle. Thank you.
 
  • #12
hunt_mat
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There's not much to learn about hyperbolic functions apart from their definitions, their graphs and their derivatives. I am not sure of their geometric interpretation, there must be one, hyperbolic geometry perhaps?

mat
 
  • #13
Char. Limit
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I just looked it up, they are supposed to give measures of a hyperbola, I believe. Somehow.
 

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