# Integrating substitution problem?

1. Dec 7, 2007

### malty

[SOLVED] Integrating substitution problem?

1. The problem statement, all variables and given/known data
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?

Last edited: Dec 7, 2007
2. Dec 7, 2007

### 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.

Last edited: Dec 7, 2007
3. Dec 7, 2007

### malty

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

Looking at the trigfunctions .. .1 min . .

Last edited: Dec 7, 2007
4. Dec 7, 2007

### rocomath

you will need to do a little more manipulation. anytime!

5. Dec 7, 2007

### malty

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 )

6. Dec 7, 2007

### 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.

Last edited: Dec 7, 2007
7. Dec 7, 2007

### malty

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) ..

8. Dec 7, 2007

### 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

9. Dec 7, 2007

### malty

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. Dec 7, 2007

### rocomath

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. Dec 7, 2007

### malty

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?

12. Dec 7, 2007

### rocomath

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

ex:

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

13. Dec 7, 2007

### malty

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)??

Last edited: Dec 7, 2007
14. Dec 7, 2007

### 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.

15. Dec 7, 2007

### malty

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

16. Dec 7, 2007

### rocomath

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. Dec 7, 2007

### malty

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:

Last edited: Dec 7, 2007
18. Dec 7, 2007

### 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

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

Last edited: Dec 7, 2007
19. Dec 7, 2007

### malty

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

20. Dec 7, 2007

### 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

$$I=uV-\int Vdu$$

Last edited: Dec 7, 2007