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This question already has an answer here:

Evaluate $$\int\frac{1}{\sqrt{4-x^2}}dx$$

I had this question on my calc exam today, and I have no clue how it's done. I was trying to factor 4-x² to see if I could see any patterns but no luck.

One thing I did notice was that $$\frac{d}{dx}(\arcsin(x)) =\frac{1}{\sqrt{1-x^2}} $$

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marked as duplicate by Martin Sleziak, Mark Bennet, user147263, Américo Tavares, Mike Pierce Jun 20 '15 at 1:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You should note that whenever you have a formula, and it differs from your question by a constant (like here 4 instead of 1), then you can simply factor it out, obtaining $$ \int\frac{1}{2}\frac{1}{\sqrt{1-(x/2)^2}}dx $$ Then you can apply the formula that you found out, and it turns out that the substitution $$u=x/2$$ would do the job.

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Your observation is apt. Let $x = 2\sin u$ for the $u$-substitution, and then you obtain

$$ \int \frac{1}{\sqrt{4 - x^2}} dx \;\; =\;\; \int \frac{2\cos u}{2\sqrt{1 - \sin^2u}} du \;\; =\;\; u+ c $$

where $c$ is constant. We then get $u = \sin^{-1}\left( \frac{x}{2}\right )$.

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Substitute $x=2\cos(\theta)$.

Note that $dx=-2\sin(\theta)d\theta$

Also use the fact that $\sin^2(\theta)+\cos^2(\theta)=1$

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  • $\begingroup$ By $sen(\theta)$ do you mean $\sin(\theta)$? $\endgroup$ – Martin Sleziak Jun 19 '15 at 16:58
  • $\begingroup$ Yes, sorry, I'm spanish speaker. $\endgroup$ – HeMan Jun 19 '15 at 18:34
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Hint Notice that

$$\frac{1}{\sqrt{4-x^{2}}}=\frac{1}{2\sqrt{1-\left( \frac{x}{2}\right) ^{2}}}$$

This will tell you the possible substitution you need to use or recognize immediately the answer.

Added. If you use the intuitive substitution $u=\frac{x}{2},du=\frac{1}{2}dx $, you get

$$\begin{aligned}\int\frac{1}{\sqrt{4-x^2}}dx&=\int\frac{1}{2\sqrt{1-\left( \frac{x}{2}\right) ^{2}}}dx=\int\frac{1}{\sqrt{1-u^{2}}}du=\arcsin(u)+C\\&=\arcsin\left(\frac{x}{2}\right)+C.\end{aligned} $$

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