Equivalent Integral for restricted values

**bugatti79** · February 12th 2011, 06:21 AM

Folks,

$<br /> \displaystyle u=\int \frac{dx}{\sqrt{2k-x^2}}=\int \frac{dx}{\sqrt{(\sqrt{2k})^2-x^2}}=sin^{-1}( \frac{x}{\sqrt{2k}})<br />$

Alpha Wolfram says that this is equivalent to

$\displaystyle tan^{-1}( \frac{x}{\sqrt{2k-x^2}})$ for restricted values of x and k.

How is this derived and for what range of values I wonder?

**DeMath** · February 12th 2011, 07:53 AM

See Inverse trigonometric functions - Wikipedia, the free encyclopedia

$\displaystyle{\arctan{\alpha}=\arcsin\frac{\alpha} {\sqrt{\alpha^2+1}}}$

**Danny** · February 12th 2011, 08:30 AM

The easiest way to visualize is to create a right-angled triangle to go with

$\theta = \sin^{-1} \dfrac{x}{\sqrt{2k}}\;\; \text{or}\;\; \sin \theta = \dfrac{x}{\sqrt{2k}}$

**Soroban** · February 12th 2011, 08:52 AM

Hello, bugatti79!

$\displaystyle u\;=\;\int \frac{dx}{\sqrt{2k-x^2}}\;=\;\int \frac{dx}{\sqrt{(\sqrt{2k})^2-x^2}}\;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{2k}}\right)$ .[1]

$\displaystyle \text{Alpha Wolfram says that this is equivalent to:}$

. . $\displaystyle \tan^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k-x^2}}\right)\:\text{ for restricted values of }x\text{ and }k.$

$\text}How is this derived and for what range of values?}$

From [1], we see that: . $\left|\dfrac{x}{\sqrt{2k}}\right| \,\le \,1 \quad\Rightarrow\quad |x| \,\le\,\sqrt{2k}$

. . That is: . $-\sqrt{2k} \:\le\: x \:\le \:\sqrt{2k}$

We also see that $\,k$ must be positive: . $k \,>\,0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{Let: }\,\theta \:=\:\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \quad\Rightarrow\quad \sin\theta \:=\:\dfrac{x}{\sqrt{2k}} \:=\:\dfrac{opp}{hyp}$

$\,\theta$ is in a right triangle with: . $opp = x,\;hyp = \sqrt{2k}$

Pythagorus says: . $adj \,=\,\sqrt{2k-x^2}$

$\text{We have: }\:\tan\theta \:=\:\dfrac{opp}{adj} \:=\:\dfrac{x}{\sqrt{2k-x^2}}$

$\text{Hence: }\:\theta \;=\;\tan^{\text{-}1}\left(\dfrac{x}{\sqrt{2k-x^2}}\right)$

$\text{Therefore: }\;\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \;=\;\tan^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k-x^2}}\right)$

**bugatti79** · February 13th 2011, 10:33 AM

Originally Posted by Soroban

Hello, bugatti79!

From [1], we see that: . $\left|\dfrac{x}{\sqrt{2k}}\right| \,\le \,1 \quad\Rightarrow\quad |x| \,\le\,\sqrt{2k}$

. . That is: . $-\sqrt{2k} \:\le\: x \:\le \:\sqrt{2k}$

We also see that $\,k$ must be positive: . $k \,>\,0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{Let: }\,\theta \:=\:\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \quad\Rightarrow\quad \sin\theta \:=\:\dfrac{x}{\sqrt{2k}} \:=\:\dfrac{opp}{hyp}$

$\,\theta$ is in a right triangle with: . $opp = x,\;hyp = \sqrt{2k}$

Pythagorus says: . $adj \,=\,\sqrt{2k-x^2}$

$\text{We have: }\:\tan\theta \:=\:\dfrac{opp}{adj} \:=\:\dfrac{x}{\sqrt{2k-x^2}}$

$\text{Hence: }\:\theta \;=\;\tan^{\text{-}1}\left(\dfrac{x}{\sqrt{2k-x^2}}\right)$

$\text{Therefore: }\;\sin^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k}}\right) \;=\;\tan^{\text{-}1}\!\left(\dfrac{x}{\sqrt{2k-x^2}}\right)$

Thanks Guys, looks good

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