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

Question

Im not asking ‘How to evaluate this integral’, I just want to know why putting while making the u-substitution $$x=\sin(u), dx=\cos(u)$$

This is weird to me because all what i know is that we put $u$ equal to $x$ times something (or whatever you want) in the integral, but here we’ve putted $x$ equal to $\sin(u)$ .

So why x and not $u$? and why specifically sine why not cosine or tangent or anything else?

The crux of u-substitution with trigonometric functions is to cancel out an ugly $\sqrt{1-x^2}$ term or $1+x^2$ term. Here, we want to use sine or cosine since$$\sin^2x+\cos^2x=1\qquad\implies\qquad\sin x=\sqrt{1-\cos^2x}$$ — Frank W, Mar 16 '21 at 19:28
@RyanG: Those are all valid points; perhaps you should turn them into an answer. — Joe, Mar 16 '21 at 20:07
Substituting $x=\cos(u)$ would also work here. The choice between sine and cosine is essentially arbitrary. — awkward, Mar 17 '21 at 13:35

zwim · Answer 1 · 2021-03-16T20:25:00.163

The substitution is often driven by the domain of definition of the function under the integration symbol, this is because to be able to conserve the value of the integral, the change of variable must be bijective.

In our case for $\sqrt{1-x^2}$ to be well defined we must have $x\in[-1,1]$

So the choice of a function with a range in $[-1,1]$ is compulsory, there are several choices, for instance $\sin(x),\cos(x),\tanh(x)$.

Theoretically all would give the same end result, but some may lead to a nice result after substitution, for instance here we have the relations $\begin{cases}1-\tanh(x)^2=\frac 1{\cosh(x)^2}\\\cos(x)^2+\sin(x)^2=1\end{cases}$

In both cases $\sqrt{1-x^2}$ will transform to $\sqrt{(something)^2}$ and will simplify, but $\sin,\cos$ will result in rather nice $\int \cos^2,-\int \sin^2$ integrals while the last change will result in a less nice $\int \frac 1{\cosh^3}$.

Although notice that in the end, if you lead calculations to the end:

the $\sin$ change will result in $\, \frac 12x\sqrt{1-x^2}+\frac 12\arcsin(x)$
the $\tanh$ change will result in $\, \frac 12x\sqrt{1-x^2}+\arctan(\frac{x+1}{\sqrt{1-x^2}})$

It may appear different, but they just differ by a constant value $\frac{\pi}4$, so the choice of the variable substitution doesn't change the result, just take the one that leads to the simpler calculation.

Here are two other examples $\displaystyle \int x\sqrt{\frac{x-1}{x+1}}\mathop{dx}\, $ and $\, \displaystyle \int \frac{x^4}{\sqrt{x^2-9}}\mathop{dx}$

Notice that for the first one, we need $x-1\ge 0\implies x\ge 1$ and for the second one $x\ge 3$ or $\frac x3\ge 1$.

Both suggest a function whose range is $[1,+\infty)$, and we know that $\cosh$ is such a function, also since it is a trig function the relation $\cosh^2-\sinh^2=1$ also leads to nice simplifications. And indeed, such a substitution allows to solve both integrals effectively.

Joe · Accepted Answer · 2021-03-16T20:30:44.053

The 'textbook answer' to your question is that making the substitution $x=\sin\theta$ allows us to apply the identity $\cos^2+\sin^2=1$. Any integral of the form $$ \int f\left(\sqrt{1-x^2}\right) \, dx \, , $$ can be attacked by making this substitution. If $x=\sin\theta$, then $\sqrt{1-x^2}=\cos\theta$ and $dx=\cos\theta d\theta$, meaning that the integral is transformed to $$ \int f(\cos\theta)\cos\theta \, d\theta \, , $$ which often makes things simpler. When I was first learning about integration by substitution, I found this explanation unconvincing for three reasons:

If $x=\sin\theta$, then $\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$. And $\sqrt{\cos^2\theta}=|{\cos\theta}|$, not $\cos\theta$.
When defining a variable, is it really correct to write 'let $x=\ldots$' as opposed to 'let $\theta=\ldots$'?
Integration by substitution is meant to be the chain rule in reverse. But here, it doesn't look like we are applying the reverse chain rule.

It is certainly possible to answer these objections, but the full answer is quite lengthy. Let me try to address them as concisely as I can—and I'll provide links if you're interested in the details:

As user170231 has mentioned in the comments, when we make the substitution $x=\sin\theta$, what we are really doing is making the substitution $\theta=\arcsin x$. This means that, by definition, $\theta$ is between $-\pi/2$ and $\pi/2$ radians. And so $\cos\theta$ is positive. In general, to integrate $$ \int f(g(x)) \, dx \, , $$ we can make the substitution $u=g(x)$, provided that $g$ is invertible. Then, we always rewrite this as $x=g^{-1}(u)$. As a shortcut, we say things like 'let $x=\sin u$', where it is understood that we are actually making the substitution $u=\arcsin x$.
I can't offer you a definitive answer to this question, but it does seem to me like it is a mild abuse of notation to define a variable by using it in an expression. However, we all know what was intended, and so this small point can be disregarded.
We are still doing the chain rule in reverse, even though it doesn't look like we are. First, if you are unfamiliar with the links between substitution and the chain rule, then see here. For more details, see here.

Regarding point (1), we want the substitution to be reversible, especially in the case of an indefinite integral. In particular $x=\sin\theta\implies\theta=\sin^{-1}x$ only for $-1\le x\le1$, and this ensures $\cos \theta\ge0$, hence $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$. — user170231, Mar 16 '21 at 19:50

score 3 · Answer 3 · edited Jul 08 '21 at 16:13

3

Note that the objective of the substitution is to simplify the integrand, i.e. removing the square-root operation. So, any substitution that achieves that is applicable, including

$$x= \sin{t},\>\>\>x=\cos{t},\>\>\>x=\tanh{t},\>\>\>x=\operatorname{sech}{t} \>\>\>etc.$$

edited Jul 08 '21 at 16:13

goldenesgeld

43

answered Mar 16 '21 at 19:47

Quanto

97,352

score 1 · Answer 4 · answered Mar 16 '21 at 19:29

Since you want to get the $x$ out of the square root, and $1-x^2$ must be a square of something.Also, $0 \le 1-x^2 \le 1$. So the something should be $\sin u$ or $\cos u$. This is a natural choice. There could be other choices, but this seems the best one so far.

score 1 · Answer 5 · answered Mar 16 '21 at 20:32

Another hint. If we know $$ \int\frac{dx}{\sqrt{1-x^2}} = \arcsin x + C \tag1$$ then it is not too hard to write your integral in terms of $(1)$. Instead of fiddling around with your integral to get $(1)$ somehow, we can just substitute $x = \sin\theta$ and let the algebra do it for us.

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

5 Answers5