# Is there a visualization for inverse trig functions as indefinite integrals

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?

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 $$\operatorname{arcsec} x + \operatorname{arccsc} x = \pi/2.$$

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:

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

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.

• The derivation is nice but your integrals have no ranges. Where in your derivation is the $\int_0^x$ part? – spraff Oct 16 '18 at 21:26
• @Spraff How about now? – clathratus Oct 16 '18 at 22:57