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I had this integral as an exercise: $\int {1\over \sqrt{a^2-x^2}}dx$.

On one hand: $\int {1\over \sqrt{a^2-x^2}}dx=\int {1\over a\sqrt{1-({x\over a})^2}}dx$. ${x\over a}=t$, $x=at$ $\Rightarrow dx=adt$ and I get $\int {1\over a\sqrt{1-t}}adt=\int {1\over \sqrt{1-t^2}}dt=\arcsin t+c=\arcsin ({x\over a})+c$.

On the other hand: $\int {1\over \sqrt{a^2-x^2}}dx=-\int -{1\over a\sqrt{1-({x\over a})^2}}dx$. ${x\over a}=t$. I get $-\int -{1\over a\sqrt{1-t}}adt=-\int -{1\over \sqrt{1-t^2}}dt=-\arccos n t+c=-\arccos ({x\over a})+c$.

Those two functions are definitely not the same. What could explain this?

I would appreciate your help.

Meitar Abarbanel
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    http://math.stackexchange.com/questions/1116974/why-its-true-arcsinx-arccosx-frac-pi2 – lab bhattacharjee Mar 13 '15 at 17:45
  • First, your two $c$'s are not required to be the same. It's more like $c_1$ and $c_2$. Second, actually your two functions are indeed the same, up to the constant difference. Looking at a unit circle and alternately interpreting $x$ as a width on the $x$-axis or a height on the $y$-axis, it becomes clear that $\arcsin(x)=\frac{\pi}{2}-\arccos(x)$. – 2'5 9'2 Mar 13 '15 at 19:51

4 Answers4

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hint: $\dfrac{d}{dx}\left(\arcsin (x) + \arccos (x)\right) = 0$.

DeepSea
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Let $\arcsin x=y\implies -\dfrac\pi2\le y\le\dfrac\pi2$

$\implies x=\sin y=\cos\left(\dfrac\pi2-y\right)\implies\arccos(x)=2m\pi\pm\left(\dfrac\pi2-y\right)$

where $m$ is an integer such that $0\le2m\pi\pm\left(\dfrac\pi2-y\right)\le\pi$

Taking $+$ sign, $-\dfrac\pi2\le -y\le\dfrac\pi2\iff0\le\dfrac\pi2-y\le\pi$ $\iff2m\pi\le2m\pi+\dfrac\pi2-y\le(2m+1)\pi$

$\implies m=0$

Consequently, $\arccos(x)=2\cdot0\pi+\left(\dfrac\pi2-y\right)=\dfrac\pi2-\arcsin(x)$

Test for $-$ sign

lab bhattacharjee
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1

$$\arcsin x+\arccos x=\frac{\pi}2$$

Daniel Fischer
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RE60K
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1

Draw the right angled triangle with angles $ \alpha,\beta, \pi/2 $ and sides $ x, \sqrt {1-x^2},1. $

$ \cos \alpha = \sin \beta $

$ \alpha + \beta = \pi/2 $

$ \cos ^{-1} x = \sin ^{-1}\sqrt{1-x^2} $

In differentiation constant $\pi/2$ vanishes.

Narasimham
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