Question

Integrate dx/sqrt(9 + x^2)

Answer 1

looks like a hyperbolic sine

Answer 2

try trig substitution

Answer 3

or this.
\[\int\limits_{?}^{?}\frac{ dx }{ \sqrt{9+x ^{2}} } \]

Answer 4

yeah

Answer 5

@satellite73 actually gave out the clue I just realized that

Answer 6

hmm. not 100 percent sure how to do this.

Answer 7

\[
9+9\tan^2(\theta)=9\sec^2(\theta)
\]

Answer 8

there might be some other ways, but quite cumbersome

Answer 9

Consider \(x^2=9\tan^2(\theta)\) and \(dx=3\sec^2(\theta)d\theta\)

Answer 10

I don't think he's familiar with trig substitution yet...

Answer 11

why does x^2 equal that though? We had went over it a little my last day of class but then we had spring break so i didnt quite grasp it.

Answer 12

http://www.math.smith.edu/Local/cicintro/ch11.pdf

Answer 13

Well, can you think of any other nice trig identity where \(1+f^2(x) = g^2(x)\)?

Answer 14

You could also try \(9 + 9\cot^2(\theta) = 9\csc^2(\theta)\) as well if you want. Then \(x^2=9\cot^2(\theta)\).

Answer 15

so... are we using trig functions just because we could set any value we wanted for x? im not quite sure what the reasoning is for this trig to come in which is why i am lost :(

Answer 16

@jeremyggg The reason for the Trig Sub is that making use of Trig Identities allows us to get rid of addition/subtraction signs between terms.

If I have something like this:\[\Large\bf\sf \sqrt{1+\tan^2 x}\]I can't take the square root easily, the addition is causing problems! :(
But if we make use of our Trig Identity,\[\Large\bf\sf \sqrt{1+\tan^2x}\quad=\quad \sqrt{\sec^2x}\]It becomes a lot more manageable!\[\Large\bf\sf =\quad \sec x\]Does that help explain it a lil bit?
Trig Identities make our integrands much easier to work with.

Answer 17

it helps but i dont understand where we are turning \[\sqrt{9+x ^{2}}\] into a trig identity?

Answer 18

All of the square identities involve 1 + (something)^2.
See how we have a 9 and not a 1?
Bad bad bad, we'll have to get it into the form 1+(something)^2.

Factoring a 9 out of each term,
\[\Large\bf\sf \sqrt{9+x^2}\quad=\quad \sqrt{9\left(1+\frac{1}{9}x^2\right)}\]We'll do some sneaky magic and bring the 1/9 into the square as (1/3)^2.
(And also bring the 9 out of the root).\[\Large\bf\sf =\quad 3\sqrt{1+\left(\frac{x}{3}\right)^2}\]

Now we have the form 1+(something)^2 under the root.
The Square Identity Involving addition is \(\Large\bf\sf 1+\tan^2\theta=\sec^2\theta\).
So we're going to make the substitution,
\[\Large\bf\sf \frac{x}{3}\quad=\quad \tan \theta\]

Answer 19

\[\Large\bf\sf 3\sqrt{1+\left(\frac{x}{3}\right)^2}\quad=\quad 3\sqrt{1+\left(\tan \theta\right)^2}\quad=\quad 3\sqrt{1+\tan^2\theta}\]We can make use of our Trig Identity from there, yes? :o

Answer 20

0.o yes!

Answer 21

hmm. so i got \[\int\limits_{?}^{?}\frac{ dx }{ \sqrt{9(1+\tan ^{2}(\theta)} }\]
and then
\[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\frac{ dx }{ \sqrt{\sec ^{2}\theta} }\]
is this the right track?

Answer 22

Yes very good.
It can simplify a bit further right?
You can take the square root of a square.

Then we need to switch out the differential for something involving \(\Large d\theta\).

Answer 23

Before we can integrate*

Answer 24

Trig Substitutions are a long process.

1. Getting everything in terms of theta is a bit of work.
2. Simplifying and then integrating isn't too bad (since you've made it easier to work with).
3. Undoing your substitution is a bit of a pain. You have to draw out a triangle and use your trig relations (Soh Cah Toa) to get your result back in terms of x.

I can explain it if you get there at some point :)
Seems like you're still working on steps 1 and 2 though for now, which is fine.

Answer 25

the dx is \[\sec ^{2}\theta d\theta\]
so im at \[\frac{ 1 }{ 3 }\int\limits_{?}^{?}\sec\theta\ d \theta \]

Answer 26

do i mess with the back sub before i integrate?

Answer 27

1/3 lncos(theta) +c