root integral question

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June 24th 2011, 02:36 PM
transgalactic

root integral question

how to solve $\int\sqrt{a-x^{2}}dx$
i tried by parts but its not not working
any ideas
June 24th 2011, 02:41 PM
Plato

Re: root integral question

Quote:

Originally Posted by transgalactic

how to solve $\int\sqrt{a-x^{2}}dx$
i tried by parts but its not not working any ideas

Look at this.
June 24th 2011, 02:48 PM
transgalactic

Re: root integral question

what method they used?
June 24th 2011, 02:50 PM
Plato

Re: root integral question

Quote:

Originally Posted by transgalactic

what method they used?

Click the button "show steps" at the right of the box.
June 24th 2011, 03:33 PM
TheCoffeeMachine

Re: root integral question

See this. If you can calculate or know

$\int\frac{1}{\sqrt{a-x^2}}\;{dx}$

then doing the integral is very easy.
June 24th 2011, 08:51 PM
chisigma

Re: root integral question

Quote:

Originally Posted by TheCoffeeMachine

See this. If you can calculate or know

$\int\frac{1}{\sqrt{a-x^2}}\;{dx}$

then doing the integral is very easy.

In any 'manual' You can find the 'derivative' ...

$\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1-x^{2}}}$ (1)

... so that is...

$\frac{d}{dx} \sin^{-1} \frac{x}{a} = \frac{1}{\sqrt{a^{2}-x^{2}}}$ (2)

On WolframAlpha [I wonder why...] a different and more complicated formula is reported...

Wolfram Mathematica Online Integrator

Anyway following TCM we obtain...

$\int \sqrt {a^{2}-x^{2}}\ dx = \frac{x}{2}\ \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\ \sin^{-1} \frac {x}{a} + c$ (3)

Kind regards

$\chi$ $\sigma$
June 25th 2011, 05:34 AM
Soroban

Re: root integral question

Hello, transgalactic!

Quote:

$\text{Integrate: }\:I \;=\;\int\sqrt{a-x^2}\:dx$

Use "Trig Substitution".
Evidently, you aren't familiar with it . . .

Let: $x = a\sin\theta \quad\Rightarrow\quad dx \,=\,a\cos\theta\,d\theta$

Note that: $\sqrt{a^2-x^2} \:=\:\sqrt{a^2-a^2\sin^2\!\theta} \:=\:\sqrt{a^2(1-\sin^2\!\theta)}$
. . . . . . . . $=\;\sqrt{a^2\cos^2\!\theta} \;=\;a\cos\theta$

Substitute: . $I \;=\;\int a\cos\theta(a\cos\theta\,d\theta) \;=\;a^2\int\cos^2\!\theta\,d\theta$

. . . . . . $=\;\frac{a^2}{2}\int(1 + \cos2\theta)\,d\theta \;=\;\frac{a^2}{2}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C$

. . . . . . $=\;\frac{a^2}{2}\left(\theta + \tfrac{1}{2}\cdot2\sin\theta\cos\theta\right) + C \;=\;\frac{a^2}{2}(\theta + \sin\theta\cos\theta) + C$ .[1]

Back-substitute: . $x \,=\,a\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{a} \:=\:\frac{opp}{hyp}$

$\theta$ is in a right triangle with: $opp = x,\;hyp = a$
Pythagorus gives us: . $adj \:=\:\sqrt{a^2-x^2}$

Hence: . $\cos\theta \:=\:\frac{\sqrt{a^2-x^2}}{a}$

We have: . $\begin{Bmatrix}\theta &=& \sin^{-1}\!\left(\dfrac{x}{a}\right) \\ \\[-3mm] \sin\theta &=& \dfrac{x}{a} \\ \\[-3mm] \cos\theta &=& \dfrac{\sqrt{a^2\!-\!x^2}}{a} \end{Bmatrix}$

Substitute into [1]:

. . $I \;=\;\frac{a^2}{2}\left[\sin^{-1}\!\left(\frac{x}{a}\right) + \frac{x}{a}\cdot\frac{\sqrt{a^2-x^2}}{a}\right] + C$

. . $I \;=\;\frac{a^2}{2}\left[\sin^{-1}\!\left(\frac{x}{a}\right) + \frac{x\sqrt{a^2-x^2}}{a^2}\right] + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I have no idea why so many textbooks and sources use $\tan^{-1}$
. . which requires extra steps and looks more intimidating.

$\text{Why not use: }\text{sec}^{-1}\left(\frac{a}{\sqrt{a^2-x^2}}\right) \;\text{ and confuse }everyone?$
June 25th 2011, 05:36 AM
HallsofIvy

Re: root integral question

Quote:

Originally Posted by transgalactic

how to solve $\int\sqrt{a-x^{2}}dx$
i tried by parts but its not not working
any ideas

Let x= a sin(t) so that dx= a cos(t) dt. $a^2- x^2= a^2- a^2 sin^2(t)= a^2(1- sin^2(t))= a^2 cos^2(t)$ so that $\sqrt{a^2- x^2}= a cos(t)$

$\int\sqrt{a^2- x^2}dx= a\int cos^2(t)dt$

And, of course, $cos^2(t)= \frac{1}{2}(1+ cos(2t))$

Just realized that this is $\int \sqrt{a- x^2}dx$ not $\int\sqrt{a^2- x^2$ .

Okay, just replace my "a" with $\sqrt{a}$ : it is
$\sqrt{a}\int cos^2(t)dt= \frac{\sqrt{a}}{2}\int (1+ cos(2t))dt$
$= \frac{\sqrt{a}}{2}(t- \frac{1}{2}sin(2t)$
$= \frac{\sqrt{a}{2}(sin^{-1}(x/\sqrt{a})- \frac{1}{2}sin(2 sin^{-1}(x/\sqrt{a}})$

$sin(2\theta)= 2 sin(\theta)cos(\theta)$ and $cos(sin^{1}(x/\sqrt{a}))= \sqrt{1- x^2/a= \frac{1}{a}\sqrt{a- x^2}$
so the integral is
$\frac{\sqrt{a}}{2}(sin^{-1}(x/\sqrt{a})- \left(\frac{x}{\sqrt{a}}\right)\left(\frac{1}{\sqr t{a}})(\sqrt{a- x^2}$
$= \frac{\sqrt{a}}{2}(sin^{-1}(x/\sqrt{a})- \frac{x}{a\sqrt{a- x^2}})$
June 25th 2011, 06:10 AM
tom@ballooncalculus

Re: root integral question

Wow!! Absolutely everyone has mis-read the a as a^2! (Rofl)

Not massively consequential, but an excuse for some pictures...

http://www.ballooncalculus.org/draw/internal/ten.png
Then, swap the inner function...
http://www.ballooncalculus.org/draw/internal/tena.png

... where (key in spoiler) ...

Spoiler:

http://www.ballooncalculus.org/asy/chain.png
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
The general drift is...
http://www.ballooncalculus.org/asy/maps/internal.png

Spoiler:

http://www.ballooncalculus.org/draw/internal/tenb.png

Spoiler:

http://www.ballooncalculus.org/draw/internal/tenc.png Larger

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
June 25th 2011, 08:12 AM
Deveno

Re: root integral question

i believe the substitution that actually works is $u = \sqrt{a}\sin(x)$