Thread: Equivalent Integral for restricted values

1. Equivalent Integral for restricted values

Folks,

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

Alpha Wolfram says that this is equivalent to

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

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

2. See Inverse trigonometric functions - Wikipedia, the free encyclopedia

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

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

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

4. Hello, bugatti79!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

5. Originally Posted by Soroban
Hello, bugatti79!

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

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

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

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

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

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

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

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

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

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

Thanks Guys, looks good