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Integrating substitution problem?

  • Thread starter malty
  • Start date
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well that's not what i got

it was ~38 something (and i also checked on my calculator) :p

and i know why you messed up, your constant is not supposed to be 5/3. if you wanna keep going, i don't mind ... i can keep talking :-]
 

malty

Gold Member
160
0
well that's not what i got

it was ~38 something (and i also checked on my calculator) :p

and i know why you messed up, your constant is not supposed to be 5/3. if you wanna keep going, i don't mind ... i can keep talking :-]
Gah this is going back to your warning now isn't it? *Totally forgot about it until now :uhh:*

*Bangs head off desk*
 
1,750
1
Gah this is going back to your warning now isn't it? *Totally forgot about it until now :uhh:*

*Bangs head off desk*
yes! hehe

so what we have here is

[tex]\frac{2}{3}[2\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta d\theta][/tex]

from here, divide by 2

[tex]\frac{2}{3}[\int_{0}^{5}\sec^{3}\theta d\theta=\frac{1}{2}\sec\theta\tan\theta|_{0}^{5}-\frac{1}{2}\int_{0}^{5}\sec\theta d\theta][/tex]

so we have solved for our I

[tex]\frac{2}{3}[\frac{1}{2}\sec\theta\tan\theta|_{0}^{5}-\frac{1}{2}\int_{0}^{5}\sec\theta d\theta][/tex]

which becomes

[tex]\frac{1}{3}[\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta d\theta][/tex]

which gives you your correct answer, see how annoying that constant can be? but you just gotta treat it right and you shouldn't have any problems with it.

here is secant to the 5th ... i was doing everything correctly as you did, but the constant kept messing me up.

http://alt1.mathlinks.ro/Forum/latexrender/pictures/2/c/6/2c677c0644a5d4317451afd0ad717909202f2be8.gif [Broken]
 
Last edited by a moderator:

malty

Gold Member
160
0
yes! hehe

so what we have here is

[tex]\frac{2}{3}[2\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta d\theta][/tex]

from here, divide by 2

[tex]\frac{2}{3}[\int_{0}^{5}\sec^{3}\theta d\theta=\frac{1}{2}\sec\theta\tan\theta|_{0}^{5}-\frac{1}{2}\int_{0}^{5}\sec\theta d\theta][/tex]

so we have solved for our I

[tex]\frac{2}{3}[\frac{1}{2}\sec\theta\tan\theta|_{0}^{5}-\frac{1}{2}\int_{0}^{5}\sec\theta d\theta][/tex]

which becomes

[tex]\frac{1}{3}[\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta d\theta][/tex]

which gives you your correct answer, see how annoying that constant can be? but you just gotta treat it right and you shouldn't have any problems with it.

here is secant to the 5th ... i was doing everything correctly as you did, but the constant kept messing me up.

http://alt1.mathlinks.ro/Forum/latexrender/pictures/2/c/6/2c677c0644a5d4317451afd0ad717909202f2be8.gif [Broken]
[/URL]

:bugeye:

Wow!!
Will have to look at that tomorrow morning . .(er technically today morning)

As for the the constant think i think I get what you did, but I don't understand why mine didn't work?

Anyways, My brain is fried now, so I'm probably wrong but shouldn't there be a half before the integral sec(theta)d(theta) and how do you type latek so darn fast!!

I bow my hat to you roco,


Thank you so much for your help, , ,my only regret is that with my exams looming so close spending 3 hours on this integral probably wasn't the wisest choice, but I just couldn't let it pass..

:approve:
 
Last edited by a moderator:
1,750
1
:bugeye:

Wow!!
Will have to look at that tomorrow morning . .(er technically today morning)

As for the the constant think i think I get what you did, but I don't understand why mine didn't work?

Anyways, My brain is fried now, so I'm probably wrong but shouldn't there be a half before the integral sec(theta)d(theta) and how do you type latek so darn fast!!

I bow my hat to you roco,


Thank you so much for your help, , ,my only regret is that with my exams looming so close spending 3 hours on this integral probably wasn't the wisest choice, but I just couldn't let it pass..

:approve:
which part are you referring to for the part i bolded?

you'll become a much better problem solving by just spending hours on one problem :-] it's not about quantity, but quality ... lol. some problems are just worth the time.

good luck on your exam! you'll do fine, just be careful!!!

also, i type in latek so fast b/c i copy/paste like crazy :-]
 
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Gib Z

Homework Helper
3,344
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Integration of Powers of Secant is not so difficult if you use a recursive formula. Theres also less chance of error until the last few lines, since we haven't substituted any numbers for the pro numerals yet. It's not difficult to derive the formula, it takes about 2 minutes.

I know rocophysics will like this challenge, so I the only hint I give to start is let to use integration by parts =]
 

Defennder

Homework Helper
2,579
5
It's quite tedious to integrate the cube of sec theta. My suggestion is that you first try to figure out how to integrate that as an indefinite integral then apply the result to evaluate the definite integral in the question. You'll definitely need to use integration by parts here, and if you do it correctly, you notice a point where you'll need to bring an integral on the RHS of the equation to to the LHS. You'll need at least 2 substitutions on this problem (just to integrate (sec(theta))^3 )
 

Gib Z

Homework Helper
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Once again someone ignores every other post in the thread >.<
 

arildno

Science Advisor
Homework Helper
Gold Member
Dearly Missed
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1. Homework Statement
Sorry to hijack this thread sorta (as a similar named one already exists), but the title is aptly suited to my question.

I have integral to integrete and I don't really know how to do it tbh. . .

[tex] s=\int{\sqrt{2+(3t)^2}dt [/tex] Limits are from t=0 to t=5 (how do I show this on latek?)

If it was 2-3t then I'd simply substitute the t for sin u, but what do I do when it positive?
I don't see why people bother with the tan substitution.

It is by far simplest to set:
[tex]t=\frac{\sqrt{2}}{3}Sinh(u)[/tex]
whereby the integrand resolves itself to [tex]\frac{2}{3}Cosh^{2}(u)=\frac{3}{4}(Cosh(2u)+1)[/tex]
 

Gib Z

Homework Helper
3,344
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Minor correction: The coefficient on the right hand side should be 1/3.
 
1,750
1
Integration of Powers of Secant is not so difficult if you use a recursive formula. Theres also less chance of error until the last few lines, since we haven't substituted any numbers for the pro numerals yet. It's not difficult to derive the formula, it takes about 2 minutes.

I know rocophysics will like this challenge, so I the only hint I give to start is let to use integration by parts =]
I'll give it a try after my finals :-]

Once again someone ignores every other post in the thread >.<
Never!!!

I don't see why people bother with the tan substitution.

It is by far simplest to set:
[tex]t=\frac{\sqrt{2}}{3}Sinh(u)[/tex]
whereby the integrand resolves itself to [tex]\frac{2}{3}Cosh^{2}(u)=\frac{3}{4}(Cosh(2u)+1)[/tex]
I actually do hyperbolic substitutions when I'm integrating, but I think it was far more beneficial to do it the long way here to learn how important the constant is and the usefulness of trig substitution.
 

malty

Gold Member
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Hyperbolic functions eh?

Well to be honest, I'd would never even have consider them, learn something new everyday I guess.

To be honest though I much prefered the Tan method because it was fun (and annoying when I think of my integration by parts loop and not spotting the obvious i.e. Integral of Sec ^2 x equal Tan x) and it was also my first time attempting a secant to the power of .. integral.

Will also give a derivation proposed by Gibz a go over the holidays, no idea where to start though, but hopefully between now and then some sort of idea will have occurred.
 
1,750
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If you really get bored and want to become a Integrating Master (lol), do some of the problems in this thread ... I've only completed 5 of them since I don't have much spare time. Some make you just want to JUMP! :-]

How Good Am I? - https://www.physicsforums.com/showthread.php?t=149706
 

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