# Intergration (1-(X^2))^0.5

EDIT: my tex is a little broken trying to fix

So i want to intergrate

$$\int (1 - x^2)^\frac{1}{2} dx$$

i start off by saying

$$\sin u = x$$

so

$$\frac {dx} {du} = \cos u$$

then

$$\int (1 - x^2)^\frac{1}{2} \cos u du$$

which is

$$\int \cos^2 u du$$

and

$$\cos2u = 2cos^2 u - 1$$

so therefore

$$\frac {1} {2} \cos2u + \frac {1} {2}= cos^2 u$$

so you intergrate

$$\int \frac {1} {2} \cos 2u + \frac {1} {2} du$$

which is

$$\frac {1} {4} \sin 2u + \frac {u} {2}$$

and

$$\sin 2u = 2\sin u \cos u = 2\sin u(1 - sin^2 u)^\frac {1} {2}$$

putting x into u gets

$$\frac {x} {2} (1 - x^2)^\frac {1} {2} + \frac {\sin^-1 x} {2}$$

which im pretty sure is wrong. So can someone show me how to intergrate (1-(X^2))^0.5 ? i think that using x = sinu is wrong but u = sinx doesnt get me far either. Probably something simple ive overlooked.

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arildno
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Absolute Values...

arildno said:
Absolute Values...
Is that a subtle hint to stick the natural log function in there? because if you say

$$u = 1 - x^2$$

then

$$\int \frac{u^\frac{1}{2}}{(4 - 4u)^\frac{1}{2}} du$$

which looks sort of f'(x)/f(x) ish... ok tbh i have no clue why you said absolute value except it means always positive and is often in log when intergrating for some reason?

arildno
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$$\sqrt{x^{2}}=???$$

EDIT: tex is a work in progress... again :\

arildno said:
$$\sqrt{x^{2}}=???$$
I'm guessing due to the fact -x and x give the same answer

$$\sqrt{x^{2}}=|x|$$

So

$$(1-(X^2))^\frac {1}{2} = (1-(|x| ^2))^\frac {1}{2}$$

$$\frac {x} {2} (1 - |x| ^2)^\frac {1} {2} + \frac {\cosec x} {2}$$

I'm worried about the fact trig has appeared in the intergral because f(x) didnt have any? And differential of cosec x is -cosecx cotx ???

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arildno
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No..according to your (correct guess we must have:
$$(\cos^{2}(u))^{\frac{1}{2}}=|\cos(u)|$$
agreed?

arildno said:
No..according to your (correct guess we must have:
$$(\cos^{2}(u))^{\frac{1}{2}}=|\cos(u)|$$
agreed?
OK... so when i change cosu to sinu... since i have

$$(\cos^{2}(u))^{\frac{1}{2}}=(1 - \sin^{2}(u))^{\frac{1}{2}}$$

then i get

$$(|1 - sin{2}(u)|)^\frac{1}{2}$$ out of it?

$$\frac {x} {2} (|1 - x^{2}|)^\frac {1} {2} + \frac {\sin^{-1}(x)} {2}$$

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arildno
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arildno said:
instead of $$\int \cos^2 u du$$

i should have

$$\int \cos(u) | \cos(u) | du$$

???

arildno
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That is correct!

Ok so how do i intergrate

$$\int \cos(u) | \cos(u) | du$$

???

btw i found a trial of a program called "Derive 6" on a demo disk i got from a maths course. It intergrated

$$\int (1 - x^2)^\frac{1}{2} dx$$

like this (click on thumbnail for bigger picture):

http://img218.imageshack.us/img218/6961/intergration1hl2.th.png [Broken]

Using that method can you avoid absolute values? I tried putting

$$\int \cos(u) | \cos(u) | du$$

into it and it didnt like it much. Got stuck if you tried to get it to do it step by step. This is what it got if you went straight to the answer.

http://img213.imageshack.us/img213/4346/intergration2xv0.th.png [Broken]

I don't think it likes or uses absolute values much. When i told it to intergrate tanx it didnt use absolute values in the answer unlike my text book. So i'm hoping intergrating |cosx| isnt hard or complicated and the programs messing up?

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arildno
You lose no generality by assumin $0\leq{u}\leq{\pi}$