Integration help

Hi,

I was hoping someone might have some clues on how to solve this integral:


sec(x)
______ dx
a - tan(x)


I can't find any useful substitutions so if anyone has any ideas I would appreciate the help.


Thanks!

first of all: sec(x) = 1/cos(x) and tan(x) = sin(x) / cos(x), so your equation is the same as:
sec(x) / (a - tan(x)) dx
= 1 / ( cos(x) (a - tan(x)) ) dx
= 1 / ( a cos(x) - cos(x) tan(x) ) dx
= 1 / ( a cos(x) - sin(x) ) dx
Maybe this helps

Anyway, according to Mathematica the solution is:
http://integrals.wolfram.com/webMath...ontsize=Medium

Hi,


Thanks for your reply. No matter which way i manipulate I still can't figure out how to arrive at the solution. I used MathCAD to arrive at a solution as well but I still have no idea how to get the solution myself. I probably should have mentioned this.

Thanks.

Did MathCAD provide the same solution as Mathematica? Knowing the solution you might be able to work backwards to discover what the right substitution is.

VBAhack

This is an interesting one. I did some looking in my old (really old) calculus book, and this problem seems to fit a class where the substitution u = tan 1/2 x might be useful. I did some playing around and got it to:

2du/(a(1-u^2) - 2u)

but couln't get any further.

VBAhack

Quote:

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Originally Posted by twanvl

Anyway, according to Mathematica the solution is:
http://www.vbforums.com/

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You can transform this solution somewhat if you don't like hyperbolic functions:

Quote:

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Originally Posted by VBAhack

This is an interesting one. I did some looking in my old (really old) calculus book, and this problem seems to fit a class where the substitution u = tan 1/2 x might be useful.
VBAhack

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This is actually the way to go. From

u = tan(x/2)

Some elementary trig gives:

x = 2 arctan(u)
dx = 2du/(1 + u²)
sin(x) = 2u/(1 + u²)
cos(x) = (1 - u²)/(1 + u²)

From this you arrive at this simpler integral:

2du / [a(1 - u²) - 2u]

Hopefully you know how to deal with integrands having polynomials in the denominator. Otherwise let me know and I'll be back to you.

Yep, you came up with the same result I did.

VBAhack

Quote:

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Originally Posted by VBAhack

Yep, you came up with the same result I did.

VBAhack

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Yeah, I noticed this afterwards. Actually, I meant to continue the whole derivation from there but was a bit in a hurry so if wy125 or you are interested I can post it.

krtxmrtz,

I think I have it. To integrate the following:

sec(x)
______ dx
a - tan(x)

First make the following substitution:

u = tan (x / 2)

This results (after some re-arranging) in:

2 du / (a - a u² - 2 u),

Next, make the following substitution:

y = u + 1/a

This yields:

(2 dy) / [a + 1/a - a y²] = (2/a) dy / (z² - y²), where z = sqrt[(1 + a²)/(a²)]

which is a common form. Thus:

∫(2/a) dy / (z² - y²) = (2/a) (1/z) arctanh (y/z) + C

substituting y = u + 1/a, we get:

= (2/a) (1/z) arctanh[ (u + 1/a) / z ] + C
= [2 / √(1 + a²)] arctanh[ a (u + 1/a) / √(1 + a²)] + C
= [2 / √(1 + a²)] arctanh[ 1 + a u) / √(1 + a²)] + C

and, finally, substituting u = tan (x / 2), we get the desired result:

= [2 / √(1 + a²)] arctanh{[1 + a tan (x / 2) ] / √(1 + a²)} + C

VBAhack :D

P.S. How'd you paste the superscript characters? Thanks for the tip on this

Quote:

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Originally Posted by VBAhack

How'd you paste the superscript characters?

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To have a text appear as superscript, write: [x]text[/x]

but instead of x write sup: I have to write it as a variable or else it will "do the job" and you won't see what's inside [ ] but the word "text" as superscript.

For a subscript, make x=sub

It's quite similar to using e.g. bold text: select some text you're composing, click on the "bold" button and see the tags.

Try it out and use "preview post" to see what it looks like before posting.

Quote:

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Originally Posted by VBAhack

krtxmrtz,

I think I have it.
Next, make the following substitution:

y = u + 1/a

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This was a clever substitution!

I took a similar but perhaps more complicated path by decomposing the integrand:

2du / (-u² - 2u +a) = (2 / a) * [1 / (u + p) - 1 / (u + q)] / (p - q)

where

p = (1 + sqrt(1 + a²))/a
q= (1 + sqrt(1 - a²))/a

so that the result is immediate:

Integral = [2/a(p - q)]*[ln(u + p) - ln(u + q)] = [2/a(p - q)]*[ln((u + p)/(u + q))] = [2/a(p - q)]*[ln((tan(x/2) + p)/(tan(x/2) + q))]

etc

Thanks for the tips on superscript - works great. I'd love to see a list of the other tags that are available....

Thanks also for your solution - I like it perhaps even better than mine because the resulting integrand is simpler.

One last tid bit (to be completely anal about this), is a derivation of ∫ dx / (z² – x²):

Let x = z tanh (z u), then

x / z = tanh (z u)
z u = arctanh (x / z)
u = (1 / z) arctanh (x / z)

Using the basic derivative: d/dx tanh (x) = sech² (x), we get

dx = z sech² (z u) z du
= z² sech² (z u) du

∫ dx / (z² - x²)
= ∫ z² sech² (z u) du / [ z² - z² tanh² (z u) ]
= ∫ sech² (z u) du / [ 1 – tanh² (z u) ]

Using the identity sech² (x) = 1 - tanh² (x), we get

= ∫ sech² (z u) du / sech² (z u)
= ∫ du
= u + C
= (1 / z) arctanh (x / z) + C

VBAhack

This is how I decomposed the integrand (which I've called R):

R = 2 / (-x² - 2x +a) = -2 / (x² + 2x -a)

x² + 2x - a = (x - p)(x - q)

where p and q are the 2 roots of the polynomial:

p = -1 + sqrt(1 + a)
q = -1 - sqrt(1 + a)

(notice there was a mistake in my calculated values for p and q in my earlier post: this does not invalidate the mothod though)

Then:

R = -2 / (x-p)(x-q) = M / (x - p) + N / (x - q)

where M and N are to be determined. This is done as follows:

-2 = M(x - q) + N(x - p) = (M + N)x - (Mq + Np)

This must hold for any value of x. hence:

M + N = 0 (there's no term for x on the left hand side)
Mq + Np = 2

Solving:

M = -2 / (p - q)
N = 2 / (p - q)

so that finally:

R = (-2 / (p - q)) / (x - p) + (2 / (p - q)) / (x - q)

Thanks, I see. You factored the quadratic equation, then used partial fractions to solve.