Integrating arcsin: Solving \int \sqrt{9-x^{2}}dx with step-by-step explanation

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SUMMARY

The integral \(\int \sqrt{9-x^{2}}dx\) can be solved using the substitution \(x = 3 \sin \theta\), which simplifies the expression significantly. The integral transforms to \(\int \frac{1}{\sqrt{9-x^{2}}}dx = \frac{1}{3} \arcsin\left(\frac{x}{3}\right)\), where the factor of \(1/3\) is accounted for by the differential \(dx\). Understanding the substitution and its implications is crucial for correctly evaluating the integral without losing factors. This discussion clarifies the importance of careful substitution in integral calculus.

PREREQUISITES
  • Understanding of integral calculus, specifically integration techniques.
  • Familiarity with trigonometric identities and substitutions.
  • Knowledge of the arcsine function and its integral form.
  • Ability to manipulate algebraic expressions and differentials.
NEXT STEPS
  • Study the process of trigonometric substitution in integrals.
  • Learn about the properties and applications of the arcsine function.
  • Practice solving integrals involving square roots and trigonometric identities.
  • Explore advanced integration techniques, such as integration by parts and substitution methods.
USEFUL FOR

Students and educators in calculus, particularly those focusing on integral calculus and trigonometric substitutions. This discussion is beneficial for anyone seeking to enhance their understanding of solving integrals involving square roots and arcsin functions.

Hannisch
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Homework Statement


The problem, from the very beginning, was:

\int \sqrt{9-x^{2}}dx

But this I have reduced to:

\int \sqrt{9-x^{2}}dx = \frac{x}{2} \sqrt{9-x^{2}} + \frac{9}{2} \int \frac{1}{\sqrt{9-x^{2}}}dx

My problem is that last integral - I get a factor of (1/3) times the correct answer and I don't know what to do - I simply can't see it.

Homework Equations



\int \frac{1}{\sqrt{1-x^{2}}}dx = arcsinx

The Attempt at a Solution



I look at it and want to "transform" my expression into something like the arcsin expression above. So I say:

\int \frac{1}{\sqrt{9-x^{2}}}dx = \int \frac{1}{3\sqrt{1-\frac{x^{2}}{9}}}dx

and from there get:

\int \frac{1}{\sqrt{9-x^{2}}}dx = \frac{1}{3}arcsin\frac{x}{3}

but you're not supposed to get that factor of 1/3 - you're not supposed to remove it? Can anyone explain to me what I'm missing? I've searched my textbook so many times now I'm about to throw it into the wall or something..
 
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\int \frac{{\rm d}u}{\sqrt{1-u^2}}=\arcsin u

but here we have

\int \frac{{\rm d}x}{\sqrt{1-\left(\frac x 3 \right)^2}}

u= \frac x 3 \Rightarrow {\rm d}x= ?
 
You need to use that relevant equation to deal with \frac{x}{3} and not just x. Be careful with that substitution!

Also, a more natural way to deal with the original integral would be to consider the substitution x=3 \sin \theta and go from there using some trig identities
 
Ohh, I think I get it now, thank you! I'd say it needs dx/3, so my 1/3 "disappears" (for lack of better wording).

Anyway, as for that last comment - you're probably right, but I don't think I've seen that type of thing being used before.
 
Hannisch said:
Anyway, as for that last comment - you're probably right, but I don't think I've seen that type of thing being used before.

No need to worry really. You will come across it sooner or later and it's good to know
 
You could also have used the trig substitution x = 3sinθ which should make the integral pretty easy to work with.
 

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