Why don't we include the minus sign in this integral?

Question

Here it is: $$\int \sqrt{16-x^2}dx$$ This is quite tricky to deal with and required ingenuity. But it's not impossible. I can let $x=4\cos{u}$, then we have $dx=-4\sin{u}du$.

In this way, $$\int\sqrt{16-x^2}dx$$ $$=\int \sqrt{16-16\cos^2{u}}(-4\sin{u})du$$ $$=-16\int \sqrt{1-\cos^2{u}}\sin{u}du$$ $$=-16\int \sqrt{\sin^2{u}}\sin{u}du$$

Here raises my question: in the next step, the solution just take $\sqrt{\sin^2{u}}\sin{u}=\sin^2{u}$ and continue. However, I don't think it's complete because $\sqrt{\sin^2{u}}\sin{u}=|\sin{u}|\sin{u}=±\sin^2{u}$

So, can anyone tell me what's the reason for only taking the plus (+) sign but not the minus (-) sign when doing this simplification?

Answer 1

Based on your suggested subsitution $x=4\cos u,$ indeed, $$\int\sqrt{16-x^2}\;\mathrm dx\\=-16\int \sqrt{\sin^2{u}}\sin{u}\;\mathrm du\\=-16\int |\sin{u}|\sin{u}\;\mathrm du.$$

Continuing: $$=-16\int (\sin{u})\sin{u}\;\mathrm du\\\text{(since $\sin u\geq0$ on the superset $[0,\pi]$ of the domain of integration)} \\=-16\int\sin^2{u}\;\mathrm du\\=\ldots.$$

Note that the (currently unspecified) limits of integration $[a,b]$ flip position on making this substitution; this explains the negative sign in front of the ‘$16$’.

Answer 2

Indeed you can use a substitution like $x = 4 \cos u$ to take the antiderivative of $\sqrt{16 - x^2}.$ I say an substitution "like" that, because $x = 4 \cos u$ is not a complete account of the substitution.

In a substitution into an antiderivative, you start with the original antiderivative, whose integrand is a function of some variable, say $x$; then for any particular value of $x$ you replace the integrand with a new integrand evaluated at a particular value of some other variable, say $u$. In this sense, $u$ is actually a function of $x,$ not the other way around, even though it is convenient (for the purpose of replacing occurrences of $x$ in the formula) to write the substitution with $x$ on one side of an equation. If you are using a non-invertible function for the substitution you may have multiple choices of the value of $u$ at which you can evaluate the new integrand for any particular value of $x$; but it has to be the same value of $u$ in all parts of the integrand, and it has to remain the same value of $u$ during the remaining evaluation, in order for the substitution to make sense. If, somewhere in the midst of evaluating the new integral, you forget which value of $u$ you had in mind when you made the substitution, you are likely to get an incorrect result.

You have been shown how to identify a particular value of $u$ for each $x$ by restricting the choice of $u$ to $u \in [0, \pi].$ In effect, this says that $u = \arccos\left(\frac14 x\right).$ This also implies that $\lvert \sin u\rvert = \sin u$ at every possible value of $u.$ It also implies that $\sin u \geq 0.$

But it is actually possible for $\sin u$ to have a negative value. You just have to set up your substitution to make this possible. An example of such a substitution is $u = -\arccos\left(\frac14 x\right).$ This is still a substitution in which $x = 4 \cos u$ and $\mathrm dx = 4\sin u\,\mathrm du$, so after substitution your integral becomes $$ I = -16 \int \sqrt{\sin^2 u} \sin u\,\mathrm du. $$ But the substitution $u = -\arccos\left(\frac14 x\right)$ implies that $u \in [-\pi,0]$ and that $\sin u \leq 0.$ So you have $\sqrt{\sin^2 u} = \lvert\sin u\rvert = -\sin u$ and $$ I = 16 \int \sin^2 u \,\mathrm du. $$

Continuing the evaluation from this point, you should eventually find that $$ I = 8u - 8 \sin u \cos u + C. $$ You now reverse the substitution using the rule $u = -\arccos\left(\frac14 x\right)$ so that $\cos u = \frac 14 x$ and (recalling that $\sin u \leq 0$ for such a value of $u$) $$ \sin u = - \sqrt{1 - \cos^u} = -\sqrt{1 - \frac1{16} x^2} = -\frac14 \sqrt{16 - x^2}. $$ The result is \begin{align} I &= -8 \arccos\left(\frac14 x\right) - 8\left(-\frac14 \sqrt{16 - x^2}\right)\left(\frac 14 x\right) + C \\ &= -8 \arccos\left(\frac x4 \right) + \frac12 x \sqrt{16 - x^2} + C , \end{align} in agreement with the result that you get from the substitution that makes $\sin u \geq 0.$

So you can have $\lvert\sin u\rvert = -\sin u$. But if at some point in the reverse substitution you forget that you are working with the case $\lvert\sin u\rvert = -\sin u$, you may incorrectly replace $u$ with $\arccos\left(\frac14 x\right)$ or replace $\sin u$ with $\frac14 \sqrt{16 - x^2}$, or both, and then you will get an incorrect result (which you can confirm by differentiation).

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Note that this answer is really just an example of the phenomenon described in a comment under the original question, $\sqrt{16-x^2}=(-1)^k\sin u$ where $u \in [k\pi,(k+1)\pi].$ You could just as easily substitute $u = 2\pi - \arccos\left(\frac14 x\right)$ and you would get the same result. It should be easy enough to see that adding any multiple of $2\pi$ to the right side of that equation results in the same values of $\sin u$ and $\cos u$ while causing the term that is linear in $u$ to add another constant to the constant $C.$ And we have already shown what happens for each choice of sign for $\pm \arccos\left(\frac14 x\right).$

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Also note that I have focused here on the evaluation of an antiderivative, that is, a transformation from one function of a single variable to a family of functions of the same single variable. I am not claiming that U-substitution in a definite integral, $$ \int_{\phi(a)}^{\phi(b)} f(t) \mathrm dt = \int_a^b f(\phi(u)) \frac{\mathrm d}{\mathrm du} \phi(u)\, \mathrm du $$ (this version is from ProofWiki) requires $\phi$ to be an invertible function. What happens in the case of a definite integral is merely that when evaluating the integral on the right-hand side, at each individual point between $a$ and $b$ you need to evaluate the integral in a way that is consistent with the substitution at that point. This is explained in answers that various people (including myself) have written here, here, here, and here.

Answer 3

Whether we see it or not, or like to see it or not the sign in front of $\sqrt{ f(x)} $ exists and is always $ \pm$.

When consistent positive sign is taken and holds good till final result is obtained, it is the usual practice, the negative sign could as well be taken and carried through till the final result.

It is implied and had been adopted and practiced as an unwritten convention imho.

The integral evaluation causes no confusion when we see it as

$$ =\mp 16\int \sqrt{\sin^2{u}}\sin{u} du$$

before further the final steps.

$$\int \sqrt{16-x^{2}}{\rm }x=\frac{x}{2}\sqrt{16-x^{2}}+8\arcsin\left(\frac{x}{4}\right)+C1$$

and $$\int \sqrt{16-x^{2}}{\rm }x=-\frac{x}{2}\sqrt{16-x^{2}}-8\arcsin\left(\frac{x}{4}\right)-C1$$

If we integrate top quarter circle in a simpler case,

$$ \int _0^a \sqrt{a^2-x^2} dx = \pi a^2/4 $$ for area above x-axis and if we take bottom quarter circle as continuation of the same curve (circle)

$$ -\int _0^a \sqrt{a^2-x^2} dx = -\frac{\pi/4}{a}$$ for area below the x-axis.