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I attempted the following integral from the 2022 MIT Integration Bee Qualifying Round:

$$\int\frac{1}{1+\sin x} + \frac{1}{1+\cos x}+ \frac{1}{1+\tan x} + \frac{1}{1+\cot x} + \frac{1}{1+\sec x} + \frac{1}{1+\csc x}dx$$


$$\int\frac{1-\sin x}{\cos^2 x} + \frac{1-\cos x}{\sin^2 x} + \frac{\cos x}{\cos x + \sin x} + \frac{\sin x}{\cos x + \sin x} + \frac{\cos x}{\cos x + 1} + \frac{\sin x}{\sin x + 1}dx$$
$$x\,+\,\int(\sec^2 x -\sec x\tan x)\,+\,(\csc^2 x - \csc x\cot x)dx + \int\frac{\cos x(1 - \cos x)}{\sin^2 x}+ \frac{\sin x(1 - \sin x)}{\cos^2 x}dx $$
$$x\,+\,\tan x - \sec x - \cot x + \csc x + \int\cot x\csc x-\cot^2 x + \tan x\sec x-\tan^2 x dx $$
$$x\,+\, \tan x - \sec x - \cot x + \csc x-\csc x+\sec x - \tan x + x + \cot x + x = \fbox{3x} $$
This is the correct answer, but I have been trying to find a shorter way to compute this integral. Is there maybe a way to combine some of the terms in the original integral to make it shorter, or this essentially the best way to do it? Thank you.

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The only fact from trigonometry we need are the reciprocal identities: $$\sec x = 1/\cos x, \quad \csc x = 1/\sin x, \quad \cot x = 1/\tan x$$

Using this, the integrand is recognized as the sum of three expressions of the form $$\frac1{1+u}+\frac1{1+u^{-1}} = \frac1{1+u}+ \frac{u}{u+1} = \frac{1+u}{1+u} = 1$$

That is, the integral is $$\int (1+1+1)\,dx = \int 3\, dx = 3x+C$$

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    $\begingroup$ As a technical limitation, do note that the algebraic simplification fills in removable singularities, so the indefinite-integral-as-antiderivative interpretation would also require us to remove all $x$-values of the form $\pi n/2$ or $3\pi/4 + \pi n$ for integral $n,$ giving a different $C$ for each interval created. $\endgroup$ Commented Jul 13, 2023 at 19:45
  • $\begingroup$ aren't those singularities of measure 0? They shouldn't affect C at all, regardless of the interval. $\endgroup$
    – Joe
    Commented Jul 14, 2023 at 6:26
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    $\begingroup$ @Joe If your end goal is to apply a version of the fundamental theorem of calculus, then I'd agree we can assign $C=0$ [almost] everywhere, but this restriction is improper when you're describing all antiderivatives of a continuous function on a disconnected domain. In such a case, measure doesn't even appear, and the "constant" of integration is only locally constant. $\endgroup$ Commented Jul 14, 2023 at 7:13

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