Integration help

[VBAhack]

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.

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[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.

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[krtxmrtz]

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

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.

[VBAhack]

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

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[krtxmrtz]

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

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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.

[VBAhack]

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

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

[krtxmrtz]

How'd you paste the superscript characters?

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.

[krtxmrtz]

krtxmrtz,

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

y = u + 1/a

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