1. ## Integration ...

How can I show that:

$\displaystyle \displaystyle \int{\frac{dx}{\sqrt{4-x^2}}=\sin^{-1}(\frac {x}{2})$

2. Trig sub 1-cos^2 = sin^2 use this fact.

3. By substitution make $\displaystyle x = 2\sin \theta$

4. Originally Posted by pickslides
By substitution make $\displaystyle x = 2\sin \theta$
$\displaystyle \displaystyle x=2\sin \theta$

$\displaystyle \displaystyle dx=2\cos\theta d\theta$

$\displaystyle \displaystyle d\theta=\frac {dx}{2\cos \theta}$

$\displaystyle \displaystyle d\theta=\frac {dx}{2\sqrt{1-\sin^2 \theta}}$

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

$\displaystyle \displaystyle d\theta=\frac {dx}{\sqrt{4-x^2}}$

$\displaystyle \displaystyle \theta=\int{\frac {dx}{\sqrt{4-x^2}}}=\sin^{-1}(\frac{x}{2})+C$

OK Thanks

5. You still haven't really "shown" it...

Draw a right-angled triangle, perpendicular height $\displaystyle x,$

hypotenuse 2 and base $\displaystyle \sqrt{4-x^2}$

such that the angle $\displaystyle \theta$ is opposite the side $\displaystyle x.$

Then...

$\displaystyle \displaystyle\ Sin\theta=\frac{x}{2}\Rightarrow\frac{d}{dx}Sin\th eta=\frac{d\theta}{dx}\;\frac{d}{d\theta}Sin\theta =\frac{1}{2}$

$\displaystyle \Rightarrow\ 2Cos\theta\;d\theta=dx$

Therefore

$\displaystyle \displaystyle\frac{1}{Cos\theta}=\frac{2}{\sqrt{4-x^2}}$

$\displaystyle \displaystyle\int{\frac{dx}{\sqrt{4-x^2}}}=\frac{1}{2}\int{\frac{1}{Cos\theta}2Cos\the ta}d\theta=\theta=Sin^{-1}\left[\frac{x}{2}\right]$

6. Originally Posted by Archie Meade
You still haven't really "shown" it...

Draw a right-angled triangle, perpendicular height $\displaystyle x,$

hypotenuse 2 and base $\displaystyle \sqrt{4-x^2}$

such that the angle $\displaystyle \theta$ is opposite the side $\displaystyle x.$

Then...

$\displaystyle \displaystyle\ Sin\theta=\frac{x}{2}\Rightarrow\frac{d}{dx}Sin\th eta=\frac{d\theta}{dx}\;\frac{d}{d\theta}Sin\theta =\frac{1}{2}$

$\displaystyle \Rightarrow\ 2Cos\theta\;d\theta=dx$

Therefore

$\displaystyle \displaystyle\frac{1}{Cos\theta}=\frac{2}{\sqrt{4-x^2}}$

$\displaystyle \displaystyle\int{\frac{dx}{\sqrt{4-x^2}}}=\frac{1}{2}\int{\frac{1}{Cos\theta}2Cos\the ta}d\theta=\theta=Sin^{-1}\left[\frac{x}{2}\right]$
$\displaystyle x=2\sin \theta$

it is clear that:

$\displaystyle \sin \theta=\frac{x}{2}$

$\displaystyle \theta=\sin^{-1}(\frac{x}{2})$

7. ## With explanation ....

$\displaystyle \displaystyle x=2\sin \theta$

$\displaystyle \displaystyle dx=2\cos\theta d\theta$

$\displaystyle \displaystyle d\theta=\frac {dx}{2\cos \theta}$

We know that:

$\displaystyle \displaystyle \sin^2 \theta +\cos^2 \theta =1$

$\displaystyle \displaystyle \cos \theta =\sqrt {1-\sin^2 \theta }$
$\displaystyle \displaystyle d\theta=\frac {dx}{2\sqrt{1-\sin^2 \theta}}$

$\displaystyle \displaystyle x=2\sin \theta$

$\displaystyle \displaystyle \sin \theta=\frac{x}{2}$

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

$\displaystyle \displaystyle d\theta=\frac {dx}{\sqrt{4-x^2}}$

$\displaystyle \displaystyle \theta=\int{\frac {dx}{\sqrt{4-x^2}}}$

$\displaystyle \displaystyle x=2\sin \theta$

$\displaystyle \displaystyle \sin \theta=\frac{x}{2}$

$\displaystyle \displaystyle \theta =\sin^{-1}(\frac{x}{2})$
$\displaystyle \displaystyle \theta=\int{\frac {dx}{\sqrt{4-x^2}}}=\sin^{-1}(\frac{x}{2})+C$