Great Scott Please Help (Trig Integration Substitution)

[hammy]

I have been trying to figure out this problem for hours. I'm probably doing something ridiculously stupid, but, alas, I truly have no clue where to go next.

PROBLEM: ?(4-x[sup:32hjfs8g]2[/sup:32hjfs8g])[sup:32hjfs8g]1/2[/sup:32hjfs8g]

One of the many ways I've attempted (this way got me closest):

?(4(1-(x[sup:32hjfs8g]2[/sup:32hjfs8g]/4)))[sup:32hjfs8g]1/2[/sup:32hjfs8g] - Take out a 4

-------------------------------------------------------------------
//BEGIN OFF TO THE SIDE WORK
(x[sup:32hjfs8g]2[/sup:32hjfs8g]/4) = sin[sup:32hjfs8g]2[/sup:32hjfs8g]?

SQUARE BOTH SIDES
(x/2) = sin? --> THEREFORE --> ? = arcsin(x/2)

x = 2sin?
dx = 2cos? d?

//END OFF TO THE SIDE WORK
-------------------------------------------------------------------

SUBSTITUTE IN NEW STUFF
?((4(1-sin[sup:32hjfs8g]2[/sup:32hjfs8g]?))[sup:32hjfs8g]1/2[/sup:32hjfs8g])(2cos?)(d?)

USE TRIG IDENTITY
?((4(cos[sup:32hjfs8g]2[/sup:32hjfs8g]?))[sup:32hjfs8g]1/2[/sup:32hjfs8g])(2cos?)(d?)

EXECUTE SQUARE ROOT FUNCTION
?(2(cos?))(2cos?)(d?)

?(4(cos[sup:32hjfs8g]2[/sup:32hjfs8g]?)(d?)

TAKE OUT 4
4?(cos[sup:32hjfs8g]2[/sup:32hjfs8g]?)(d?)

USE TRIG IDENTITY
4?(1/2)(1+cos2?)(d?)

TAKE OUT (1/2)
2?(1+cos2?)(d?)

DISTRIBUTE
2?(d?)+(cos2?)(d?)

SEPARATE INTO TWO PARTS
2?(d?) + 2?(cos2?)(d?)

INTEGRATE
2? + sin2?

RESUB in (?=arsin(x/2))?
sin2? + 2arcsin(x/2)

And I have no clue how to get from there, to the answer given here:
2arcsin(x/2) + ((1/2)(x))(((4-x[sup:32hjfs8g]2[/sup:32hjfs8g]))[sup:32hjfs8g]1/2[/sup:32hjfs8g]) + C

 

[galactus]

Use the sub \(\displaystyle x=2sin\theta\)

After all is done, resub \(\displaystyle \theta=sin^{-1}(x/2)\) to get it in terms of x again.

\(\displaystyle sin(2(sin^{-1}(x/2)))+2sin^{-1}(x/2)\)

\(\displaystyle sin(2(sin^{-1}(x/2)))=\frac{x\sqrt{4-x^{2}}}{2}\)

 

[hammy]

Thanks so much for the help, I think I just don't understand the final trig step you're taking there, going from sin(2(arcsin(x/2))) = (x((3-x[sup:2tigxx74]2[/sup:2tigxx74]))^1/2)/2

I only learned arcsin rules once about 3 years ago, can you explain what is going on there in that substitution?

 

[galactus]

This has to do with inverse trig functions. You probably encountered them not long before the trig sub.

I will show it algebraically, but it can be shown with right triangles.

We have \(\displaystyle sin(2sin^{-1}(x/2))\)

Let \(\displaystyle \theta=sin^{-1}(x/2)\)

So, \(\displaystyle sin(2\theta)=2sin\theta cos\theta\)

resub:

\(\displaystyle 2sin(sin^{-1}(x/2))cos(sin^{-1}(x/2))\)

Thus, \(\displaystyle sin(2sin^{-1}(x/2))=2(x/2)(\frac{\sqrt{4-x^{2}}}{2})\)

 

[hammy]

Omg. I feel so stupid. Please... feel free to insult me; all of you great math gods. I am not worthy of your greatness. Thanks so much.