# Proving the integral $\int{1\over \sqrt{1-x^2}}\,dx$ via substitution

$$\int{1\over \sqrt{1-x^2}}\,dx$$

How do I solve this without $$\sin$$ or $$\cos$$ substitution? I want to try using $$u$$ substitution because I feel like it will be more intuitive for me. So far, I have tried to make the function to the power of a negative exponent:

$${(\sqrt{1-x^2})}^{-1}$$

I tried to do $$u$$ $$=$$ $$1-x^2$$, and do the substitution method, but I keep getting the wrong answer; I'm trying to use this indefinite integral to solve a definite one from $$1/2$$ to $$\sqrt{3}$$/$$2$$.

I'd rather not use trigonometric substitution because I want to do this assuming I don't already know that this is the derivative of $$\sin^{-1}x$$.

• Have you tried using integration by parts? – Toby Mak Mar 15 '20 at 23:31
• I don't see how you can get around using trigonometric functions since the integral's solution is one too. – Jam Mar 15 '20 at 23:31
• If you use $u = 1-x^2,$, then $du = -2x \,dx$. So to replace dx, you need $dx= \frac {du}{-2x}.$ But then you need to define $-2x$ in terms of $u = 1-x^2$. Your best route is to use $x = \sin \theta$ or $x = \cos \theta.$ Say we pick $x= \sin \theta$. Then $dx = \cos \theta.$ And we have the integral $\int \frac{\cos \theta}{\sqrt{\cos \theta}}\,d\theta$ – amWhy Mar 15 '20 at 23:31
• Sajjib, because we can then take advantage of the identity $\cos^2 \theta = 1-\sin^2 \theta$, or $\sin^2 \theta = 1-\cos^2 \theta$. – amWhy Mar 15 '20 at 23:39
• The link provided by @MaximilianJanisch is awesome: It explains and develops the most frequently used trig substitutions. I'll repeat the link here. Most basically, it amounts to knowing a few trig identities. – amWhy Mar 16 '20 at 0:40

Let $$y(x)$$ be defined on $$(-1,1)$$ as

$$y(x)=\int_0^x \frac1{\sqrt{1-t^2}}\,dt\tag1$$

Note that $$y(0)=0$$.

Differentiating $$(1)$$, we find that

$$\frac{dy}{dx}=\frac1{\sqrt{1-x^2}}$$

whence we see that

$$\frac{dx}{dy}=\sqrt{1-x^2}\tag2$$

Note that $$x'(0)=1$$.

Differentiating $$(2)$$ reveals

\begin{align} \frac{d^2x}{dy^2}&=-\frac{x}{\sqrt{1-x^2}}\,\frac{dx}{dy}\\\\ &=-x\tag3 \end{align}

The general solution to the ODE of $$(3)$$ is $$x(y)=A\sin(y)+B\cos(y)$$. Given that $$x(0)=0$$ and $$x'(0)=1$$, we find that $$x(y)=\sin(y)$$. This implies that $$y(x)$$ is the inverse function of the sine function.

Denoting this inverse function $$y(x)=\arcsin(x)$$ yields the result

$$\int_0^x \frac1{\sqrt{1-t^2}}\,dt=\arcsin(x)$$

• +1 elegant answer Mark – Satyendra Mar 16 '20 at 1:11
• @LostInSpace Thank you! Much appreciate your note! – Mark Viola Mar 16 '20 at 1:15

Try the self similar substitution $$x=\frac{1-t}{1+t},\quad t\ge 0$$ $$\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}=-\int{\frac{1}{\sqrt{t}\left( 1+t \right)}dt}=-2\int{\frac{1}{\left( 1+{{u}^{2}} \right)}du}$$

It is not really possible to avoid a trigonometry-less solution, as the antiderivative is $$\arcsin$$ and nothing else and is not expressible as a rational function (though it has a logarithmic expression in the imaginary numbers).

The nice solution by @logo is right, but strictly speaking it should also require a substitution such as $$u=\tan \theta$$.

In fact,

$$\frac1{\sqrt{1-x^2}}$$ should be considered an elementary integral as it appears in a table of derivatives of the common functions, just like, say, $$\dfrac1x$$.

Using the inverse function theorem, you know that the antiderivative will be a solution of the ODE

$$y'=\sqrt{1-y^2}.$$

This should ring the bell of the sine function, for which we know that $$(\sin y)'=\cos y$$ and $$\cos y=\sqrt{1-\sin^2y}$$.

Then,

$$x=\sin y\iff y=\sin^{-1}x.$$