Confusion about the definition of indefinite integral.

Question

let $f$ be a function and $\frac{d}{dx} F= f$. We know $\int f(x) dx=\{F(x)+c\colon c\in \mathbb R\}$, that is, it is a set it is not one element. Also, we if $A$ is a set and $A+A=\{a+b\colon a,b\in A\}$ so $A+A\neq 2A.$ Consider the following problem $A:=\int e^x \sin(x) dx$, so we have $$A=-e^x \cos(x)+e^x\sin(x)-A$$ and $$2A= -e^x \cos(x)+e^x\sin(x).$$ I have seen this solution in many books, maybe you do so. Based of my understanding of the indefinite integral as a set so the above integral is not correct. Any thought?

Answer 1

The fact that $\int e^x\sin x\,dx + \int e^{x}\sin x\,dx = 2\int e^x\sin x\,dx$ is a consequence of the linearity of the integral. $$\begin{align*} \int e^x\sin x\,dx + \int e^x\sin x\,dx &= \int (e^x\sin x + e^x\sin x)\,dx\\ &= \int 2e^x\sin x\,dx,\\ &= 2\int e^x\sin x\,dx. \end{align*}$$ Even if you want to interpret the indefinite integral as a set, you still get the right "answer": the functions are of the form $F(x)+C$ with $F$ fixed and $C$ a constant, and we if we add a function of the form $F(x)+C$ with a function $F(x)+D$ wee get $2F(x)+(C+D)$; that is, the elements of $A+A$ really are "take the fixed function $F(x)$, multiply it by two, and add a constant", i.e., $2A$.