Integrating substitution problem?

In summary: Not sure if this is right the order, but my past experiences with these buggers is that the exponents are usually rarely let equal u if there is an alegrbaic or trig function present .. am I right?correct! make sure you keep writing your integrand on the right side = to your workex:\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta = \mbox{work...}\Ok, just before I start can I just confirm this is the sensible thing to do:Yes, that's correct.
  • #1
malty
Gold Member
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[SOLVED] Integrating substitution problem?

Homework Statement


Sorry to hijack this thread sort of (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?
 
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  • #2
you want to get the radican in the form of a+bt^2, where b=1

[tex]s=\int_{0}^{5}\sqrt{2+9t^{2}}dt[/tex]

to show your limits on your integrand ... \int_{0}^{5}

so

[tex]s=3\int_{0}^{5}\sqrt{\frac{2}{9}+t^{2}}dt[/tex]

since your last problem was negative, you were able to use sine. this is negative, so what trigonometric identities do you know? options are secant and tangent.
 
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  • #3
rocophysics said:
you want to get the radican in the form of a+bt^2, where b=1

[tex]s=\int_{0}^{5}\sqrt{2+9t^{2}}dt[/tex]

to show your limits on your integrand ... \int_{0}^{5}

so

[tex]s=3\int_{0}^{5}\sqrt{\frac{2}{9}+t^{2}}dt[/tex]

Ahh nice... (still stuck) thanks a million and for the latek help too:D

Looking at the trigfunctions .. .1 min . .
 
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  • #4
malty said:
Ahh nice... (still stuck) thanks a million and for the latek help too:D

Looking at the trigfunctions .. .1 min . .
you will need to do a little more manipulation. anytime!
 
  • #5
malty said:
Ahh nice... (still stuck) thanks a million and for the latek help too:D

Looking at the trigfunctions .. .1 min . .

Okkk I'm still stuck here..., if i take either sec or tan i'll get positives and I think i want to change it to a minus to the form 1-cos^2 (t) or sin the form. Or do I need to get this ... (you this is what happens when you develop a reliance upon the same thing working each time :redface:)
 
  • #6
well between secant or tangent, you want tangent b/c it gives you a positive in the radican

[tex]\tan^{2}\theta+1=\sec^{2}\theta[/tex]

the only thing you have to do is get it in the form of 1+tan^2(x)

a little algebra

[tex]s=3\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}+t^{2}}dt[/tex]

still the same, so next let ...

[tex]t=\frac{\sqrt{2}}{3}\tan{\theta}[/tex]

now take the derivative and substitute where it is appropriate.
 
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  • #7
rocophysics said:
well between secant or tangent, you want tangent b/c it gives you a positive in the radican

[tex]\tan^{2}x+1=\sec^{2}x[/tex]

the only thing you have to do is get it in the form of 1+tan^2(x)

a little algebra

[tex]s=3\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}+t^{2}}dt[/tex]

still the same, so next let ...

[tex]t=\frac{\sqrt{2}}{3}\tan{\theta}[/tex]

now take the derivative and substitute where it is appropriate.

Thanks very much for sticking with me on this :)

i've got it down to here [tex] \sqrt{2}\int_{t=0}^{t=5}\sqrt{\frac{2}{9}(1+\tan^2 \theta)}\sec^2\theta \hspace{4} d\theta [/tex] ...
hope that is right...
ok so the sec will cancel giving sec^3 I'm assuming i just let [tex] u=\sec {\theta} [/tex]

.. (still on it) ..
 
  • #8
you have everything correct so far except the constant inside the radical, it should be

[tex]s=\sqrt{2}\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}[1+\tan^{2}\theta]}\sec^{2}\theta d\theta[/tex]

which becomes

[tex]s=\frac{2}{3}\int_{0}^{5}\sqrt{1+\tan^{2}\theta}\sec^{2}\theta d\theta[/tex]

from there, you cannot just use a simple u=substitution, you will need to use Integration by Parts
 
  • #9
rocophysics said:
you have everything correct so far except the constant inside the radical, it should be

[tex]s=\sqrt{2}\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}[1+\tan^{2}\theta]}\sec^{2}\theta d\theta[/tex]

which becomes

[tex]s=\frac{2}{3}\int_{0}^{5}\sqrt{1+\tan^{2}\theta}\sec^{2}\theta d\theta[/tex]

from there, you cannot just use a simple u=substitution, you will need to use Integration by Parts


I was wondering what on Earth I should do to get the sec and tan out of there!
Ahh ok... woah this is cool integral.. *thanks the heavens this isn't his exam*
 
  • #10
malty said:
I was wondering what on Earth I should do to get the sec and tan out of there!
Ahh ok... woah this is cool integral.. *thanks the heavens this isn't his exam*
lol, it shocked me too! i was like no way, wtf? haha, i really hate integrating secants, such a pain, but I'm with you to the end!
 
  • #11
rocophysics said:
lol, it shocked me too! i was like no way, wtf? haha, i really hate integrating secants, such a pain, but I'm with you to the end!

Ok, just before I start can I just confirm this is the sensible thing to do:

I'm going to let u=sec (theta) and dv = sec^2 (theta) d(theta)

not sure if this is right the order, but my past experiences with these buggers is that the exponents are usually rarely let equal u if there is an alegrbaic or trig function present .. am I right?
 
  • #12
correct! make sure you keep writing your integrand on the right side = to your work

ex:

[tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta = \mbox{work done}[/tex]
 
  • #13
rocophysics said:
correct! make sure you keep writing your integrand on the right side = to your work

ex:

[tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta = \mbox{work done}[/tex]

THis is going to be a long typing . ..

Um, I'm getting an integral that seems to be going on to infinity, have done 3 int by parts and I'm stuck now with the integral of ln sec.sec^2 d(theta) .. . .wait!... nah still stuck

[tex] let \hspace {4} u=\sec {\theta} [/tex] and [tex]dv=\sec^2 {\theta} d\theta}[/tex]

This gives

[tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta =(\sec {\theta})*(\int{\sec^2 {\theta} d\theta}) - . .. . . . [/tex]

Need to work out integral of \sec^2 {\theta} d\theta.

Let [tex] k=\sec {\theta} [/tex] ... .wait should i let sec^2 theta equal 1/cos^2 (theta)??
 
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  • #14
you only had to do parts once. after the first IbP, you use a trig substitution which gives you your original integrand and from there you just take the integral of secant.
 
  • #15
rocophysics said:
you only had to do parts once. after the first IbP, you use a trig substitution which gives you your original integrand and from there you just take the integral of secant.

Urgh ...well initially I did four of them .. .:redface:
 
  • #16
malty said:
Urgh ...well initially I did four of them .. .:redface:
is all good :-]

i was once given secant to the 5th to integrate and it took me a month. little mistakes and blind vision, i feel your pain.
 
  • #17
malty said:
Urgh ...well initially I did four of them .. .:redface:

Gahhh... this thing still ain't working out.. I cannot see any resemblance from the left to the right side, i keep getting the integral in terms of ln |sec| .

[tex]\int_{t=0}^{t=5}\sec^3 {\theta} d\theta= \sec{\theta}\int{\sec^2\theta}d{\theta} .. . [/tex]

Done this ,v part, integral numerous ways now,
let sec^2 =1 +tan 2

didn't work

tried doing it by parts again and again
latest effort

let sec^2 = 1/cos^2
and when i integrated that i got ln|sec (theta):yuck:
 
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  • #18
[tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta\tan^{2}\theta d\theta[/tex]

now use a trig identity to change tan^2(x) into secant and you will get your original integrand.

now at this step, be careful! leave out your fraction of 2/3 for the next few steps. this is something i learned from secant to the 5th and kicked my ass all over the place.

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

now add secant cubed to the left side and you will get

[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]
 
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  • #19
rocophysics said:
[tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta\tan^{2}\theta d\theta[/tex]

now use a trig identity to change tan^2(x) into secant and you will get your original integrand.

now at this type, be careful! leave out your fraction of 2/3 for the next few steps. this is something i learned from secant to the 5th and kicked my ass all over the place.

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

em how did you get v = tan (theta)

When I did it I had u=sec (theta)

meaning [tex]v= \int{\sec^2{\theta} d\{theta}}[/tex] This kept giving me the ln|sec| or and integral that went on forever, .. 1 sec.. .
would you mind check to see where I'm going wrong
 
  • #20
your substitutions were

[tex]u=\sec\theta[/tex]
[tex]du=\sec\theta\tan\theta d\theta[/tex]

i think this is where you messed up

[tex]dV=\sec^{2}\theta d\theta[/tex]
[tex]V=\tan\theta[/tex]

dV also means

[tex]dV=\int\sec^{2}\theta d\theta[/tex]
[tex]V=\tan\theta[/tex]

so

[tex]I=uV-\int Vdu[/tex]
 
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  • #21
rocophysics said:
your substitutions were

[tex]u=\sec\theta[/tex]
[tex]du=\sec\theta\tan\theta d\theta[/tex]

i think this is where you messed up

[tex]dV=\sec^{2}\theta d\theta[/tex]
[tex]V=\tan\theta[/tex]

Oh right, this may sound trivial but I could never get sec^2 integrated = tan x .. i just kept getting loops :cry:
 
  • #22
You know the derivative of tan is sec2. So it's just the fundamental theorem of calculus to go the other way
 
  • #23
Office_Shredder said:
You know the derivative of tan is sec2. So it's just the fundamental theorem of calculus to go the other way

Arghhhhhhhhhhhhhhhhhhhhhhhhhhhhh


That was just ... grr.. .. i kept seeing the integral thing on the math tables (hence the ln's)... never looked at the differentiation things on the tables ...grrr... that's just the icing on the cake ..:mad:

So simple!
So bloody simple gah .. .how come I couldn't see that ... (lol i even googled abstract trigonometric identies and integrals) .. .

Ok .. .give me a few more mins at this.. *must calm down first*
 
  • #24
so how's the integration going? lol
 
  • #25
rocophysics said:
so how's the integration going? lol

.. . ..

[tex] I=\frac{2}{3} \int_{t=0}^{t5}sec^3(\theta) d\theta[/tex]

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

[tex] \frac {5}{3}I=\sec\theta\tan\theta|_{t=0}^{t=5}+\ln{\sec(\theta)+\tan(\theta)}|_{t=0}^{t=5} [/tex] .. . .

. . .
[tex] t= \frac {\sqrt{2}}{3}\tan {\theta} [/tex]

==> [tex]\theta= tan^-1{\frac{3t}{\sqrt{2}} [/tex]

@t=0 tan0 =0

@t=5 tan^-1 (15/sqrt2) =84º


*sigh*

evaluated the new limits, and got the total arc length to be 52.84, and I think I'm going to accept it.. now for the fun part as I've no answer to compare to I can't tell if it right but anyways, with the exception of evaluating the limits is the rest ok . .

Thanks a million for the help Roco I really really appreciate it:D
 
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  • #26
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 want to keep going, i don't mind ... i can keep talking :-]
 
  • #27
rocophysics said:
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 want to 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*
 
  • #28
malty said:
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 got to 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]
 
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  • #29
rocophysics said:
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 got to 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:
 
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  • #30
malty said:
: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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  • #31
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 =]
 
  • #32
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 )
 
  • #33
Once again someone ignores every other post in the thread >.<
 
  • #34
malty said:

Homework Statement


Sorry to hijack this thread sort of (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]
 
  • #35
Minor correction: The coefficient on the right hand side should be 1/3.
 

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