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Consider the integral

$$\int_a^bx\, dx$$

where $a=\sin x$, and $b=\cos x$.

How can we evaluate this particular integral, if $a$ and $b$ are both functions of $x$, which is the variable with respect to which we are integrating?

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6 Answers 6

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The variable inside the integral is a "dummy" in that it could be replaced by any other symbol. I think you could interpret $$\int_{\sin x}^{\cos x} x \, dx$$ as a sloppy way of writing, but having the same meaning as, $$\int_{\sin x}^{\cos x} y \, dy.$$

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Here is another interpretation.

Formally if $b\geq a$ then $\int_a^bxdx$ is a notation for $\int_{-\infty}^{\infty}\mathbf1_{(a,b]}(x)xdx$.

Applying that here leads to: $$\int_{\sin x}^{\cos x}xdx=\int_{-\infty}^{\infty}\mathbf1_{(\sin x,\cos x]}(x)xdx$$

I do not dare to say that this is the correct way of interpreting, but it illustrates at least that your question is a good question.

Personally I go for the interpretation of Umberto.

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    $\begingroup$ +1, this makes sense to the "incorrect" notation, so it seems fine for the case $\endgroup$
    – Masacroso
    Oct 26, 2018 at 18:34
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    $\begingroup$ +1 This is almost certainly not what is intended, but I like it :) $\endgroup$
    – Carmeister
    Oct 26, 2018 at 19:33
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    $\begingroup$ If someone is asking this question, they're at the level of introductory calculus. Confusing them with analysis will not be helpful. It is a near-certainty that they need the "dummy variable" answer. $\endgroup$
    – JKreft
    Oct 26, 2018 at 20:02
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    $\begingroup$ @manooooh the notation $\mathbf 1_A$ represent the indicator function of the set $A$. $\endgroup$
    – Masacroso
    Oct 27, 2018 at 3:10
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    $\begingroup$ @manooooh well, boolean functions and indicator functions share the same image set $\{0,1\}$. I dont know if they are related in any other way $\endgroup$
    – Masacroso
    Oct 27, 2018 at 4:53
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The particular expression \begin{align*} \int_{\sin x}^{\cos x}x\,dx\tag{1} \end{align*} is not a well-defined integral, since (1) is not a valid expression.

  • We have inside the integral the integration variable $x$, indicated by $dx$. This kind of variable is called a bound variable, similarly as the index $i$ in $\sum_{i=0}^n i$.

  • On the other hand $x$ is also used in the upper and lower limit of the integral, i.e. it is used as free variable, similarly as the variable $n$ in $\sum_{i=0}^ni$.

Since a variable can be only either bound or free within its scope (i.e. the range of validity of the variable) the expression (1) is not valid.

Hint: Based on experience we sometimes identify typos and are inclined to correct them with something meaningful. Nevertheless we have to be careful when doing so and when we are in doubt, we shouldn't do anything which can't be justified by mathematical rules.

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Alternatively to Umberto P. you could also view it as:

$\int_A f(x) dx$ where $A$ is a set, for example an area or volume, that is further distorted by your function $f(x)$.

In your case you would simply plug in your constraints: $A = \{a \in \mathbb{R} | \sin(x) \le a \le \cos(x) \}$ and for $f(x)$ you have of course: $f(x) = x$

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    $\begingroup$ If someone is asking this question, they're at the level of introductory calculus. Confusing them with analysis will not be helpful. It is a near-certainty that they need the "dummy variable" answer. $\endgroup$
    – JKreft
    Oct 26, 2018 at 20:02
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Doesn't matter. Just your final solution will also be a function of x.

Proceed the usual way: find the indefinite integral $$\int xdx=\frac{x^2}{2}$$ put limits: $a=\sin(x)$ and $b=\cos(x)$ $$=\frac{\cos^2x}{2}-\frac{\sin^2x}{2}$$ $$=\frac{\cos(2x)}{2}$$

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With regards to actually evaluating the integral, I think that I would interpret it the same as drhab, that:

$$\int_{\sin x}^{\cos x}xdx=\int\mathbf1_{(\sin x,\cos x]}(x)xdx$$

and given that:

$$\int x\ dx \implies f(x) = F'(x) = x$$

we can then do (where $C$ is the constant of integration):

$$\int_{\sin x}^{\cos x} x \, dx = \frac{x^2}{2} + C$$

and then:

$$\frac{x^2}{2} \bigm|_{sin\ x}^{cos\ x}\ =\ \frac{(cos\ x)^2}{2}\ -\ \frac{(sin\ x)^2}{2}\ =\ \frac{(cos\ x)^2-(sin\ x)^2}{2}$$

finally, simplify:

$$\frac{(cos\ x)^2-(sin\ x)^2}{2}=\frac{cos\ 2x}{2}$$

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