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Integrating substitution problem?

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malty

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[SOLVED] Integrating substitution problem?

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?
 
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1,750
1
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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malty

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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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1,750
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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!
 

malty

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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:)
 
1,750
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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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malty

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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) ..
 
1,750
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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
 

malty

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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*
 
1,750
1
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!
 

malty

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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 gonna 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?
 
1,750
1
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]
 

malty

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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 gonna 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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1,750
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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.
 

malty

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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:
 
1,750
1
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.
 

malty

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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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1,750
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[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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malty

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[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
 
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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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malty

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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:
 

Office_Shredder

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You know the derivative of tan is sec2. So it's just the fundamental theorem of calculus to go the other way
 

malty

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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*
 
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so how's the integration going? lol
 

malty

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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 gonna 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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