# Integrating substitution problem?

#### malty

Gold Member
[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. . .

$$s=\int{\sqrt{2+(3t)^2}dt$$ 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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#### rocomath

you want to get the radican in the form of a+bt^2, where b=1

$$s=\int_{0}^{5}\sqrt{2+9t^{2}}dt$$

so

$$s=3\int_{0}^{5}\sqrt{\frac{2}{9}+t^{2}}dt$$

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

Gold Member
you want to get the radican in the form of a+bt^2, where b=1

$$s=\int_{0}^{5}\sqrt{2+9t^{2}}dt$$

so

$$s=3\int_{0}^{5}\sqrt{\frac{2}{9}+t^{2}}dt$$
Ahh nice... (still stuck) thanks a million and for the latek help too:D

Looking at the trigfunctions .. .1 min . .

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#### rocomath

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

Gold Member
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 )

#### rocomath

well between secant or tangent, you want tangent b/c it gives you a positive in the radican

$$\tan^{2}\theta+1=\sec^{2}\theta$$

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

a little algebra

$$s=3\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}+t^{2}}dt$$

still the same, so next let ...

$$t=\frac{\sqrt{2}}{3}\tan{\theta}$$

now take the derivative and substitute where it is appropriate.

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#### malty

Gold Member
well between secant or tangent, you want tangent b/c it gives you a positive in the radican

$$\tan^{2}x+1=\sec^{2}x$$

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

a little algebra

$$s=3\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}+t^{2}}dt$$

still the same, so next let ...

$$t=\frac{\sqrt{2}}{3}\tan{\theta}$$

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 $$\sqrt{2}\int_{t=0}^{t=5}\sqrt{\frac{2}{9}(1+\tan^2 \theta)}\sec^2\theta \hspace{4} d\theta$$ ...
hope that is right...
ok so the sec will cancel giving sec^3 I'm assuming i just let $$u=\sec {\theta}$$

.. (still on it) ..

#### rocomath

you have everything correct so far except the constant inside the radical, it should be

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

which becomes

$$s=\frac{2}{3}\int_{0}^{5}\sqrt{1+\tan^{2}\theta}\sec^{2}\theta d\theta$$

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

#### malty

Gold Member
you have everything correct so far except the constant inside the radical, it should be

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

which becomes

$$s=\frac{2}{3}\int_{0}^{5}\sqrt{1+\tan^{2}\theta}\sec^{2}\theta d\theta$$

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*

#### rocomath

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

Gold Member
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?

#### rocomath

correct! make sure you keep writing your integrand on the right side = to your work

ex:

$$\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta = \mbox{work done}$$

#### malty

Gold Member
correct! make sure you keep writing your integrand on the right side = to your work

ex:

$$\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta = \mbox{work done}$$
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

$$let \hspace {4} u=\sec {\theta}$$ and $$dv=\sec^2 {\theta} d\theta}$$

This gives

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

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

Let $$k=\sec {\theta}$$ ... .wait should i let sec^2 theta equal 1/cos^2 (theta)??

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#### rocomath

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

Gold Member
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 .. .

#### rocomath

Urgh ...well initially I did four of them .. .
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

Gold Member
Urgh ...well initially I did four of them .. .
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| .

$$\int_{t=0}^{t=5}\sec^3 {\theta} d\theta= \sec{\theta}\int{\sec^2\theta}d{\theta} .. .$$

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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#### rocomath

$$\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta\tan^{2}\theta d\theta$$

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.

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

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

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

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#### malty

Gold Member
$$\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta\tan^{2}\theta d\theta$$

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.

$$\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec^{3}\theta+\int_{0}^{5}\sec\theta$$
em how did you get v = tan (theta)

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

meaning $$v= \int{\sec^2{\theta} d\{theta}}$$ 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

#### rocomath

$$u=\sec\theta$$
$$du=\sec\theta\tan\theta d\theta$$

i think this is where you messed up

$$dV=\sec^{2}\theta d\theta$$
$$V=\tan\theta$$

dV also means

$$dV=\int\sec^{2}\theta d\theta$$
$$V=\tan\theta$$

so

$$I=uV-\int Vdu$$

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#### malty

Gold Member

$$u=\sec\theta$$
$$du=\sec\theta\tan\theta d\theta$$

i think this is where you messed up

$$dV=\sec^{2}\theta d\theta$$
$$V=\tan\theta$$
Oh right, this may sound trivial but I could never get sec^2 integrated = tan x .. i just kept getting loops

#### Office_Shredder

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

#### malty

Gold Member
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 ..

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*

#### rocomath

so how's the integration going? lol

#### malty

Gold Member
so how's the integration going? lol
.. . ..

$$I=\frac{2}{3} \int_{t=0}^{t5}sec^3(\theta) d\theta$$

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

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

. . .
$$t= \frac {\sqrt{2}}{3}\tan {\theta}$$

==> $$\theta= tan^-1{\frac{3t}{\sqrt{2}}$$

@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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