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I was solving this exercise, however it is proposed that there is a possible contradiction in the exercise but I cannot determine what it is.

The idea is to find the integral of $\int{\frac{1}{\sin(x)\cos(x)}dx}$ For this purpose, the following is expressed

$\int{\frac{1}{\sin(x)\cos(x)}dx}=\int{\frac{\cot(x)}{\cos^2(x)}dx}=\int{\cot(x)\tan'(x)dx}=\cot(x)\tan(x)-\int{\tan(x)\cot'(x)dx}=1+\int{\frac{\tan(x)}{\sin^2(x)}dx}=1+\int{\frac{1}{\sin(x)\cos(x)}dx}$

Where does the failure occur? thank you for your help.

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2 Answers 2

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There is no failure. You've shown that $$\int{\frac{1}{\sin(x)\cos(x)}dx}=1+\int{\frac{1}{\sin(x)\cos(x)}dx},$$ which is correct, even if unhelpful: recall that $\int f(x)\,dx$ denotes the set of all antiderivatives of $f(x)$, all of which are a constant apart from one another. This means that it is in general true that for any function $f(x)$ we have $\int f(x)\,dx = 1 + \int f(x)\,dx$. Your derivation is correct, but it does not lead to a solution of the integral.

(To actually solve the integral, you can use the Weierstrass half-angle substitution, i.e. $t=\tan(\frac{x}{2})$, and you may simplify beforehand using $\sin(2x)=2\sin(x)\cos(x)$.)

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  • $\begingroup$ I was consulting and an interesting idea came to my mind, it is clear that when carrying out an integration by parts a constant must be added, in this case we have the constant $1$, when it should be the constant $1+C$; but I am not very sure about that. $\endgroup$
    – Wrloord
    Mar 16, 2022 at 3:56
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    $\begingroup$ @BAYRONIGNASIOLEONCIPRIAN The point is that the symbol $\int f(x)\,dx$ already includes in it all constants, because it means "the set of all antiderivatives of $f(x)$". When we write $\int f(x)\,dx = F(x)+C$ the right hand side has to have the $+C$ added, but when both sides have the $\int$ sign then the meaning of $\int f(x)\, dx = \int g(x)\, dx$ is that the set of all antiderivatives of $f(x)$ is the set of all antiderivatives of $g(x)$, and so there is no need for the constant. Therefore you can write $\int x\, dx = 1 + \int x\, dx$ and also $\int x\, dx = 80 + \int x\, dx$ etc. $\endgroup$
    – Snaw
    Mar 16, 2022 at 4:08
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HINT

Another way to solve the proposed integral: \begin{align*} \int\frac{\mathrm{d}x}{\sin(x)\cos(x)} & = 2\int\frac{\mathrm{d}x}{\sin(2x)}\\\\ & = 2\int\frac{\sin(2x)}{\sin^{2}(2x)}\mathrm{d}x\\\\ & = -\int\frac{\mathrm{d}(\cos(2x))}{1 - \cos^{2}(2x)}\\\\ & = \int\frac{\mathrm{d}(\cos(2x))}{\cos^{2}(2x) - 1} \end{align*}

Can you take it from here?

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