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February 14th, 2017, 03:21 AM   #1
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Can someone show me the quickest and easiest way to integrate this..?

Can someone show me the quickest and easiest way to integrate this..?

$\displaystyle \int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx$

Please leave your step by step instructions below. Thankyou very much indeed!!
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February 14th, 2017, 04:23 AM   #2
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The fastest way would be to reference a table of integrals. Look at #8 in the link

Table


Failing that, I'd use the trig substitution $x=a\sin{t} \implies dx= a\cos{t} \, dt$

$\displaystyle \int \sqrt{a*2 - a^2\sin^2{t}} \cdot a\cos{t} \, dt$

$\displaystyle a \int \sqrt{a^2(1-\sin^2{t})} \cos{t} \, dt$

$\displaystyle a^2 \int \cos^2{t} \, dt$

use the trig identity $\cos^2{t} = \dfrac{1+\cos(2t)}{2}$

$\displaystyle \dfrac{a^2}{2} \int 1+\cos(2t) \, dt$

$\dfrac{a^2}{2}\bigg[t + \dfrac{\sin(2t)}{2}\bigg]+C$

$x=a\sin{t} \implies t = \arcsin\left(\dfrac{x}{a}\right) \text{ and } \cos{t} = \dfrac{\sqrt{a^2-x^2}}{a}$

therefore, $\sin(2t)=2\sin{t}\cos{t}=2 \cdot \dfrac{x}{a} \cdot \dfrac{\sqrt{a^2-x^2}}{a} = \dfrac{2x\sqrt{a^2-x^2}}{a^2}$

substitution back into the antiderivative ...

$\dfrac{a^2}{2}\bigg[\arcsin\left(\dfrac{x}{a}\right) + \dfrac{x\sqrt{a^2-x^2}}{a^2} \bigg]+C$

$\dfrac{a^2}{2}\arcsin\left(\dfrac{x}{a}\right)+ \dfrac{x\sqrt{a^2-x^2}}{2}+C$
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February 14th, 2017, 05:43 AM   #3
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I think you complicated the final bit... But nevertheless, you provided the answer I was looking for. Many thanks.
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February 14th, 2017, 06:38 AM   #4
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Quote:
Originally Posted by skeeter View Post
The fastest way would be to reference a table of integrals. Look at #8 in the link

Table


Failing that, I'd use the trig substitution $x=a\sin{t} \implies dx= a\cos{t} \, dt$

$\displaystyle \int \sqrt{\color{red}{a^2}-a^2\sin^2{t}} \cdot a\cos{t} \, dt$
corrected typo ...

Quote:
I think you complicated the final bit ...
So, how would you "uncomplicate" it? Always willing to learn new integration techniques.
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February 19th, 2017, 02:15 AM   #5
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February 19th, 2017, 03:34 AM   #6
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@Soumyadip
Sorry, but I've seen Egyptian hieroglyphs that are easier to read. If someone else can read it then fine but otherwise could you write it out more slowly? It looks to me like your writing is bad due to how fast you are writing.

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February 19th, 2017, 05:32 AM   #7
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It's an approach that works, but whatever method one uses to evaluate $\displaystyle \int \frac{1}{\sqrt{a^2-x^2}}\,\mathrm dx$ (table of integrals, trigonometric substitution) works equally well for the original integral, so you've essentially just added a step.
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February 19th, 2017, 07:51 AM   #8
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OK I'm sry for my bad handwriting
It's in my habit.

U can integrate the function easily by parts
Taking sqrt(a^2 - x^2) as the 1st part and 1 as the 2nd one..

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