Question

integrate the double integral of the square root of (x^2 + 2) dydx from 0 to 1, from 0 to 1

Answer 1

\[\int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{x^2+2} dydx\]

Answer 2

so whats the problem?

Answer 3

not quite sure how to do it.. do i use u sub? o.O u = x^2 du = 2xdx 1/2du = xdx?

Answer 4

I suggest trigonometric substitution.

Answer 5

that's my kryptonite lol

Answer 6

I'll give you a hint: use x = ((2)^.5) sec(theta) as your substitution.

Answer 7

i don't even know what to do with that.. nor how you decided to use it :(

Answer 8

secant? I would have thought tangent. tan^2+1=sec^2

Answer 9

yeah it is tangent. sorry about that.
ok, here is how you do it in some detail; first you integrate with respect to y, right? After doing so you're only going to get:\[y \sqrt{x^2+1}\] from the y value 1 to 0: Thus the equation becomes \[\int\limits_{0}^{1} \sqrt{x^2+1}\]
using x = \[\sqrt{2} \tan \theta d \theta\]

Answer 10

Not done yet; accidental send. Continuing on:

Answer 11

Basically if you substitute in x=tan u, dx=sec^2 u du, you then end up with the integral of sec^3 u, which is easiest to just already know. Also some constants with the sqrt(2) and whatnot.

Answer 12

for reference, the integral of sec^3 u is 1/2[sec u tan u + log (sec u + tan u)], which you can get by integration by parts.

Answer 13

The above equation then becomes \[\int\limits_{}^{} \sqrt{2\sec^2\theta+2} (\sqrt{2} \sec \theta \tan \theta d theta)\]
since with the substitution dx = \[\sec \theta \tan \theta d \theta\].
But \[\sqrt{2 \sec^2 \theta + 2} \] by identity is \[\sqrt{2} \tan \theta\]. So the whole integral is now :\[\int\limits_{}^{}2 \sec \theta \tan^2 \theta\]. \[\tan^2 \theta = \sec^2 \theta -1\], so the above integral is now \[\int\limits_{}^{}2 \sec^3 \theta- \sec \theta\]. Now integrate everything and then substitute x back in.

Answer 14

how do you know when to use trig sub?

Answer 15

It takes practice to recognize when trig sub will be helpful. Often when you have terms like these: \(\sqrt{x^2+1}, \sqrt{x^2-1}, \sqrt{1-x^2}\).

Answer 16

do you always set x = root2 tan theta?

Answer 17

@dominusscholae You did that whole post with x=sec theta, not x = tan theta. sec^2+1 doesn't equal tan^2 lol

Answer 18

i'm so confused now :(

Answer 19

Nope, not always root2 tan theta. You use whichever one is appropriate for the situation. In this case you want tangent, because tan^2 + 1 = sec^2. You want the one that will simplify the integral for you. Give me a minute and I'll go into more detail.

Answer 20

AAAAAHHHH what a terrible mistake! Sorry I am doing all of my work on the page instead on actually on pen and paper!

Answer 21

Alright, let's do an example or two. These are the questions I used to learn how trig sub worked. First, let's look at one that doesn't actually require trig sub, but that will give us a feel for it.
\[
\int\frac{dx}{\sqrt{1-x^2}}
\]
We already recognize this as arcsin, but let's go ahead and do it with trig sub.
\[
x=\sin u\\dx=\cos u\\
\int\frac{dx}{\sqrt{1-x^2}}=\int\frac{\cos u\ du}{\sqrt{1-\sin^2u}}\\
\text{Since }(1-\sin^2u)=\cos^2u,\\
\int\frac{\cos u \ du}{\sqrt{\cos^2u}}=\int\frac{\cos u}{\cos u}du=\int du=u\\
\text{Since }x=\sin u, u=\arcsin x.
\]
So, we obtain the result that we were already expecting, by doing a little bit more work than necessary. But, that should demonstrate how trig sub works out pretty well.

Answer 22

wait.. how do we recognize it as arcsin? /blush

Answer 23

If you don't recognize it as arcsin then trig sub is actually useful there! :) Usually you would recognize it just from having derived arcsin before, since you kind of have to memorize the formulas for deriving most of the inverse trig functions. Writing up another example now, give me one moment.

Answer 24

Let's look at another example.
\[
\int \frac{dx}{\sqrt{1+x^2}}
\]
We notice that we have an \(x^2+1\), which is a decent indication that tangent could be helpful. Let's try it out:
\[
x=\tan u\\
dx=\sec^2 u\\
\int \frac{dx}{\sqrt{1+x^2}}=\int \frac{\sec^2u\ du}{\sqrt{1+\tan^2x}}=\int\frac{\sec^2u}{\sec u}du=\int\sec u\ du=\log(\sec u + \tan u)\\
\log(\sec u + \tan u) = \log(\sqrt{1+x^2}+x)
\]
The integral of secant is one that you probably just want to memorize. That last step is also a bit tricky, substituting the x back in for u. Basically, you have to keep your trig identities in mind. Substituting x for tan u is easy, because we originally set x=tan u. For the sec u, we remember that \(\sec x = \sqrt{1+\tan^2x}\) because \(\tan^2x+1=\sec^2x\).

Answer 25

you're like.. blowing my mind right now lol

Answer 26

Haha it's kind of a lot to get used to, it's best to practice it with a bunch of problems.

Answer 27

why.. when you sub the x back in .. is it +x?

Answer 28

OH nvm its cuz tan u = x

Answer 29

Yep, exactly.

Answer 30

i honestly don't know if i can wrap my mind around this lol

Answer 31

It just takes practice :)

Answer 32

thankfully, it's only a very tiny portion of the class, and most likely won't come up on the test -_-" and then, two weeks from now, i'm done with math forever ^.^

Answer 33

Being familiar with trig identities is also obviously essential. For the first while, just have a list of all the trig identities, trig derivatives, and trig integrals, and just refer to it and look for familiar parts.

Answer 34

Being done with math forever sounds terribly depressing :-/

Answer 35

lol oh on the contrary.. math and i .. do not get along.

Answer 36

Probably just because you haven't hung out enough ;) The schools tend to do a really terrible job of teaching what math is. What they teach in schools is just computation, it's like if you had a music class and they only let you play scales and never any actual pieces, let alone actually ever compose.

Answer 37

yeah maybe.. and i've had some realllyyyy bad math teachers :( all the do is copy examples from the book.. that's no way to learn lol

Answer 38

wait so, i'm still confused on this integral lol
\[\int\limits_{0}^{1} \sqrt{x^2+2}\]

Answer 39

i looked up all the trig identities, and none of them seem to fit :(

Answer 40

Okay so on this one, you want to set \(x = \sqrt{2}\tan u\). That will make \(dx=\sec^2u\). So, \[
\int_0^1\sqrt{x^2+2}\ dx=\int_0^1\sqrt{2(\tan^2u+1)}\cdot\sec^2u\ du=\sqrt{2}\int_0^1\sqrt{\sec^2u}\cdot\sec^2u\ du\\
=\sqrt{2}\int_0^1\sec^3u\ du
\]
Have you integrated \(\int\sec^3x\ dx\) before?

Answer 41

presumably, in calc 2 when i was learning trig identities.. two years ago haha

Answer 42

Haha. Well, it's one that comes up pretty often, I just always refer back to what it is instead of calculating it out. It's not too hard to calculate using integration by parts, though, and might be a good exercise to refresh some memory on trig identities. Anyway, it evaluates to \(\frac{1}{2}[\sec u\tan u+\log(\sec u +\tan u)]\). Once you have that, it's just a matter of substituting back in with x.

Answer 43

ohhh okay
awesome!! thank you soooo much!!!

Answer 44

i has another question, should i close this one and ask a new one? :o

Answer 45

Yep, that's preferred, that way people will know there's a new question :) I have to run and get some food, I'll be back in 15-20 minutes and if you still need help I'll check in.

Answer 46

okay for sure! thank you :)