Question

find the integral of 1/square root of 4-x^2 dx

Answer 1

you are supposed to make a trig sub, but you can do it without, if you recall that
\[\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\]

Answer 2

\[\int\limits \frac{ 1 }{ \sqrt{a^2-x^2}}dx=\sin^{-1} \left( \frac{ x }{a } \right)\]

Answer 3

or just go right to the answer

Answer 4

I dont get it. I'm a visual person, I need a step by step.

Answer 5

@surjithayer gave you the final form, something that is probably in the back cover of your text

Answer 6

or you can rewrite
\[\frac{1}{\sqrt{4-x^2}}\] as
\[\frac{1}{2}\frac{1}{\sqrt{1-(\frac{x}{2})^2}}\]

Answer 7

the make the substitution \(u=\frac{x}{2}\) and you get it right away

Answer 8

Whenever you see \(\sqrt{1-ax^2}\), \(\sqrt{1+ax^2}\), or \(\sqrt{ax^2-1}\) you should immediately think trig sub.

Answer 9

or you can even make \(x=2\sin(\theta)\) and so \(dx=2\cos(\theta)d\theta\) get
\[\int \frac{2\cos(\theta)d\theta}{\sqrt{4-4\sin^2(\theta)}}\]

Answer 10

Then remember \[
1-\sin^2(x) = \cos^2(x)
\]\[
1+\tan^2(x) = \sec^2(x)
\]\[
\sec^2(x)-1 = \tan^2(x)
\]

Answer 11

the denominator is \(2\cos(\theta)\) so you are integrating \(\int 1 d\theta=\theta=\sin^{-1}(\frac{x}{2})\)