# Thread: root integral question

1. ## root integral question

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

2. ## Re: root integral question

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.

3. ## Re: root integral question

what method they used?

4. ## Re: root integral question

Originally Posted by transgalactic
what method they used?
Click the button "show steps" at the right of the box.

5. ## 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.

6. ## Re: root integral question

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

$\chi$ $\sigma$

7. ## Re: root integral question

Hello, transgalactic!

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

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.

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

8. ## Re: root integral question

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

9. ## Re: root integral question

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

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

Then, swap the inner function...

... where (key in spoiler) ...
Spoiler:

... 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...

Spoiler:

Spoiler:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

10. ## Re: root integral question

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