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

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 $\sin^{-1}x$.

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    $\begingroup$ Have you tried using integration by parts? $\endgroup$ – Toby Mak Mar 15 '20 at 23:31
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    $\begingroup$ I don't see how you can get around using trigonometric functions since the integral's solution is one too. $\endgroup$ – Jam Mar 15 '20 at 23:31
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    $\begingroup$ 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 $\int \frac{\cos \theta}{\sqrt{\cos \theta}}\,d\theta$ $\endgroup$ – amWhy Mar 15 '20 at 23:31
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    $\begingroup$ Sajjib, because we can then take advantage of the identity $\cos^2 \theta = 1-\sin^2 \theta$, or $\sin^2 \theta = 1-\cos^2 \theta$. $\endgroup$ – amWhy Mar 15 '20 at 23:39
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    $\begingroup$ 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. $\endgroup$ – amWhy Mar 16 '20 at 0:40
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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

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

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  • $\begingroup$ +1 elegant answer Mark $\endgroup$ – Satyendra Mar 16 '20 at 1:11
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    $\begingroup$ @LostInSpace Thank you! Much appreciate your note! $\endgroup$ – Mark Viola Mar 16 '20 at 1:15
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Try the self similar substitution $$x=\frac{1-t}{1+t},\quad t\ge 0$$ $$\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}=-\int{\frac{1}{\sqrt{t}\left( 1+t \right)}dt}=-2\int{\frac{1}{\left( 1+{{u}^{2}} \right)}du}$$

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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, $\dfrac1x$.

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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,

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

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