Integrate $\frac{1}{\sqrt{4-x^2}}$

Question

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 $$\frac{d}{dx}(\arcsin(x)) =\frac{1}{\sqrt{1-x^2}} $$

Answer 1

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.

Answer 2

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 $u = \sin^{-1}\left( \frac{x}{2}\right )$.

Answer 3

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

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

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