Integration help

[wy125]

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!

[twanvl]

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:

[wy125]

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.

[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.

VBAhack

[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

[krtxmrtz]

Anyway, according to Mathematica the solution is:

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

[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.

VBAhack

[krtxmrtz]

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

VBAhack

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

[VBAhack]

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

[krtxmrtz]

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)

[VBAhack]

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