Is there a visualization for inverse trig functions as indefinite integrals

Question

Examining the indefinite integral formulations of inverse trig functions I notice some things

$$\arcsin(x)=\int_0^x \frac{1}{\sqrt{1-z^2}}dz$$

$$\arccos(x)=\int_x^1 \frac{1}{\sqrt{1-z^2}}dz$$

We can say that these functions "split" the range of integration $[0..1]$ at $x$.

Is there a visualization which expresses this relationship graphically?

I mean, other than just drawing the graph, is there a visualization which meaningfully shows the relationship, in an insightful and intuitive way?

Similarly,

$$\arctan(x)=\int_0^x\frac{1}{z^2+1}dz$$

$$\mathrm{arccot}(x)=\int_x^\infty\frac{1}{z^2+1}dz$$

Again, these "split" the range $[0..\infty]$ at $x$. Is there a digram for this?

Similarly,

$$\mathrm{arcsec}(x)=\int_1^x\frac{1}{z\sqrt{z^2-1}}dz$$

$$\mathrm{arccsc}(x)=\int_x^\infty\frac{1}{z\sqrt{z^2-1}}dz$$

These split the range $[1..\infty]$ at $x$. Is there a diagram for this?

Answer 1

Here is a "visual" explanation for $ \arccos $ and $ \arcsin $.

Use $ \gamma(t) = (1-t,\sqrt{1-(1-t)^2}) $ where $ 0 \le t \le 1 $ to parameterize the upper right part of the unit circle. (You can find this parameterization by a simple application of Pythagoras Theorem.) Then fix $ x = \cos(\alpha) $ where $ 0 \le \alpha \le \frac{\pi}{2} $. Now calculate $ \alpha $ in Radian by

$$ \alpha(x) = L(\gamma | _ {[0,1-x]}) = \int_{0}^{1-x} {|\gamma'(s)|ds} = \int_{0}^{1-x} {\sqrt{1 + \left(\frac{-(1-s)}{\sqrt{1-(1-s)^2}}\right)^2}} ds = \int_{0}^{1-x} {\sqrt{\frac{1-(1-s)^2+(1-s)²}{1-(1-s)^2}}} ds = \int_{0}^{1-x} {\sqrt{\frac{1}{1-(1-s)^2}} ds} = \int_{1}^{x} {\frac{-1}{\sqrt{1-s^2}} ds} = \int_{x}^{1} {\frac{1}{\sqrt{1-s^2}} ds} $$

But the function which assign's to $ x $ the value $ \alpha $ such that $ x = \cos(\alpha) $ is by definition $ \arccos $. Hence we have derived

$$ \arccos(x) = \int_{x}^{1} {\frac{1}{\sqrt{1-s^2}} ds} $$

So the integral is interpreted as measuring the length of the unitcircle from $ (1,0) $ to $ (\cos(\alpha),\sin(\alpha)) $. This is exactly the angle $ \alpha $ in radian.

Now we use $ \sin(\alpha) = \cos(\beta) $ where $ \beta = \frac{\pi}{2} - \alpha $. The angle $ \beta $ can be found in the triangle spanned by $ (0,0) $, $ (\cos(\alpha),\sin(\alpha)) $ and $ (0,1) $. As above we calculate the angle in radian by

$$ \beta(x) = L(\gamma|_{[1-x,1]}) = \cdots = \int_{1-x}^{1} { \sqrt{\frac{1}{1-(1-s)^2}} ds } = \int_{x}^{0} { \frac{-1}{\sqrt{1-s^2}} ds } = \int_{0}^{x} { \frac{1}{\sqrt{1-s^2}} ds } $$

You can interpret this integral as measuring the length of the unitcircle from $ (\cos(\alpha),\sin(\alpha)) $ to $ (0,1) $. This is exactly the angle needed to complete $ \alpha $ to $ \frac{\pi}{2} $.

So the reason of $ [0,1] $ beeing split at x is the relation $ \sin(\alpha) = \cos(\beta) $ where $ \beta = \frac{\pi}{2} - \alpha $ in combination with the analytic definition of an angle.

Note that we have also "proofed" the following result:

$$ \frac{\pi}{2} = \alpha(x) + \beta(x) = \int_0^1{\frac{1}{\sqrt{1-t^2}}dt} $$

Answer 2

Suppose the length of the hypotenuse of a right triangle is $1$ and the length of one leg is $x.$ Then the angle opposite the side of length $x$ is $\arcsin x,$ and the angle between that side and the hypotenuse is $\arccos x.$ Since the sum of the two small angles of a right triangle is $\pi/2,$ we have the identity $\arcsin x + \arccos x = \pi/2.$ The same kind of argument shows that $\arctan x + \operatorname{arccot} x = \pi/2$ and $\operatorname{arcsec} x + \operatorname{arccsc} x = \pi/2.$

Answer 3

For me, the visual has always been the derivation process. $$I=\int_0^x\frac{dt}{(1-t^2)^{1/2}}$$ Suppose $t=\sin u$, then $dt=\cos u\,du$ $$I=\int_{t=0}^{t=x}\frac{\cos u}{(1-\sin^2u)^{1/2}}du$$ $$I=\int_{t=0}^{t=x}\frac{\cos u}{(\cos^2u)^{1/2}}du$$ $$I=\int_{t=0}^{t=x}\frac{\cos u}{\cos u}du$$ $$I=\int_{t=0}^{t=x} du$$ $$I=u|_{t=0}^{t=x}$$ $$I=(\arcsin t)|_{0}^{x}$$ $$I=\arcsin x$$ I don't know man. Something about watching those integrals go from nasty to beautiful just really did the trick for me.