# root integral question

• Jun 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
• Jun 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.
• Jun 24th 2011, 02:48 PM
transgalactic
Re: root integral question
what method they used?
• Jun 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.
• Jun 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.
• Jun 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

$\chi$ $\sigma$
• Jun 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.

$\text{Why not use: }\text{sec}^{-1}\left(\frac{a}{\sqrt{a^2-x^2}}\right) \;\text{ and confuse }everyone?$

• Jun 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
$\frac{\sqrt{a}}{2}(sin^{-1}(x/\sqrt{a})- \left(\frac{x}{\sqrt{a}}\right)\left(\frac{1}{\sqr t{a}})(\sqrt{a- x^2}$
$= \frac{\sqrt{a}}{2}(sin^{-1}(x/\sqrt{a})- \frac{x}{a\sqrt{a- x^2}})$
• Jun 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

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• Jun 25th 2011, 08:12 AM
Deveno
Re: root integral question
i believe the substitution that actually works is $u = \sqrt{a}\sin(x)$