Question

Calculus 2,


problem:
https://www.dropbox.com/s/eczkps3nd3a0yez/Screenshot%202013-10-26%2022.31.59.jpg



I am stuck here:


https://www.dropbox.com/s/s66gb68im93ibob/Screenshot%202013-10-26%2022.32.09.jpg



Now i think it wants something called like "trigonometric substitution...."

Can you tell me how I can do this ?

Answer 1

Understand the steps I made there?
The whole reason we go through that mess is to GET RID OF THE SUBTRACTION!
See how taking advantage of our trig identities allowed us to simplify it down to a single term that doesn't involve addition/subtraction?

It means we can take the square root much easier.

Answer 2

So under the root we have:\[\Large 4-x^2\quad=\quad 2^2-x^2\]So we'll make the substitution:\[\Large x\quad=\quad 2\sin \theta\]

\[\large 2^2-(x)^2\quad=\quad 2^2-(\qquad)^2\]Understand where we're going with this? :o
No?

Answer 3

lol im such a doofus.. i plugged in a cos in the first example :) oh goodness....

Answer 4

Grr I gotta erase that :( It's just too big of a mistake, it's gonna confuse you...

Answer 5

\[\Large a^2-x^2\]Making the substitution:\(\Large x=a\sin\theta\)
\[\large a^2-(x)^2\quad=\quad a^2-(a \sin \theta)^2\quad=\quad a^2-a^2\sin^2\theta\quad=\quad a^2(1-\sin^2\theta)\]\[\Large=\quad a^2\cos^2\theta\]Grr :c

Answer 6

Where you at girl? :U Am I just confusing you more? lol

Answer 7

yes bro I understand so far @zepdrix I am sorry from being away I am really anxious on trying to find the answer lol

Answer 8

I have done this, I actually managed to get to a point where I don't even know how to continue anymore https://www.dropbox.com/s/s66gb68im93ibob/Screenshot%202013-10-26%2022.32.09.jpg

Answer 9

Hmm seems like you should be getting:\[\Large 2u+\sin2u+C\]Lemme double check and make sure I didn't make a silly mistake somewhere.

Answer 10

yea yea thats what I got indeed

Answer 11

So you made the initial substitution:\[\Large x\quad=\quad 2\sin u\]yes?

Answer 12

yes

Answer 13

Ok now it gets a little tricky :) We have to go back to triangle math from our trig class.
If we `solve` for sin u, we get,\[\Large \sin u\quad=\quad \frac{x}{2}\]

Answer 14

We want to draw a triangle which corresponds to this relationship.\[\Large \sin u\quad=\quad \frac{x}{2}\quad=\quad \frac{opposite}{hypotenuse}\]

Answer 15

how did u got with that formula

Answer 16

What formula? :U

Answer 17

https://www.dropbox.com/s/domccblablyo3zy/Screenshot%202013-10-26%2022.57.20.jpg

Answer 18

\[\Large 2\sin u\quad=\quad x\]Dividing each side by 2 gives us:\[\Large \sin u\quad=\quad \frac{x}{2}\]

Answer 19

Then we need to remember our triangle relations.
Sine is opposite over hypotenuse, yes?

`SOH` CAH TOA

Answer 20

hmm unfortunately I am not good at trigonometry ? :D

Answer 21

opposite over hypo is sin(x) ?

Answer 22

yes you silly billy -_- gotta remember that stuff lol

Answer 23

:/

Answer 24

how about this https://www.dropbox.com/s/zy2c8wk71ogjbo8/Screenshot%202013-10-26%2023.00.29.jpg

Answer 25

It's just a way to help us remember the relationships. You were not taught that way maybe? :o

`soh` ~ sine = (opposite)/(hypotenuse)
`cah` ~ cosine = (adjacent)/(hypotenuse)
`toa` ~ tangent = (opposite)/(adjacent)

Those are really important relationships to remember :c

Answer 26

ohh nice trick, I will use this

Answer 27

\[\Large \sin u\quad=\quad \frac{x}{2}\quad=\quad \frac{opp}{hyp}\]So I've labeled the sides of our triangle corresponding to sine of u.

Answer 28

ok

Answer 29

So what is the missing side of our triangle? :)
Use the `Pythagorean Theorem`

Answer 30

x^2 + 4 = something^2

Answer 31

so something = sqrt(x^2 + 4) ?

Answer 32

Woops you've got it setup a little backwards.

x^2 + something^2 = 2^2

The sums of the squares should equal the `hypotenuse` squared.

Answer 33

oh

Answer 34

so something = sqrt(2^2 - x^2)

Answer 35

...:D

Answer 36

So that is our missing side? Ok good.