Question

how does integral of 4dx/(sqrt(4-x^2) turn into 4arcsin(x/2)?

Answer 1

Are you familiar with Trigonometric Substitutions?

Answer 2

\[\Large\rm \int\limits \frac{4dx}{\sqrt{4-x^2}}\]

Answer 3

no i'm not

Answer 4

Hmm

Answer 5

i understand that 1/sqrt(1-x^2 turns into arcsin

Answer 6

I guess another option would be to do a little algebra, make a u-substitution, and then relate it to the integral which produces arcsine from a table or something.

Answer 7

You do? Ok great.
Let's turn it into that then :)

Answer 8

ok!

Answer 9

Inside the root, let's pull a 4 out of each term,\[\Large\rm \int\limits\limits \frac{4dx}{\sqrt{4\left(1-\frac{x^2}{4}\right)}}\]This is going to get a little messy, so try to follow along.
We want to bring the 4 into the square with the x.
4 = 2^2, yes?
So it comes in as a 2.\[\Large\rm \int\limits\limits\limits \frac{4dx}{\sqrt{4\left(1-\left[\frac{x}{2}\right]^2\right)}}\]

Answer 10

hmm ok

Answer 11

Let's pull the 4 out of the root as a 2, and simplify the 2 on bottom and 4 on top.\[\Large\rm 2\int\limits \frac{dx}{\sqrt{1-\left[\frac{x}{2}\right]^2}}\]

Answer 12

ok got that

Answer 13

We are trying to get the form that you mentioned.
So we'll make the substitution \(\Large\rm u=\dfrac{x}{2}\)

Answer 14

What do you get for the differential?
\(\Large\rm du=?\)

Answer 15

ahhh interesting!

Answer 16

you get 1/2

Answer 17

\[\Large\rm du=\frac{1}{2}dx\]Ok great!

Answer 18

Oh let's multiply the 1/2 to the other side,\[\Large\rm 2du=dx\]And then just plug everything in.
Should work out really nicely from here.

Answer 19

so you just multiply another 2 outside

Answer 20

\[\Large\rm 2\int\limits\limits \frac{2du}{\sqrt{1-\left[u\right]^2}}\]Yes, pull the new 2 outside with the other one, giving us a 4.

Answer 21

Thank you so much!

Answer 22

As a side note, we could have initially made the substitution \(\Large\rm x=2u\).
That would have avoided some of the algebra sets.
I wasn't quite sure which method would make more sense to you though.

No problem! :)

Answer 23

algebra steps*

Answer 24

how do u make the substitution of x=2u?

Answer 25

The reason to do it that way is to try and match the 4's up.
\(\Large\rm x=2u\)
Plugging it in gives us:\[\Large\rm \int\limits\limits \frac{4dx}{\sqrt{4-(2u)^2}}=\int\limits\frac{4dx}{\sqrt{4-4u^2}}\]And from there you can factor out the 4 and get that nice form that we got using the other method.
You still have to replace the dx, but maybe this way is a little simpler not pulling the 4 into the square and all that.

I dunno, you can decide for yourself which way makes more sense.

Answer 26

that is interesting as well, thank you for your tips!