# 2022 MIT Integration Bee, Qualifying Round, Question 17

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.

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 $$\int (1+1+1)\,dx = \int 3\, dx = 3x+C$$
• 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. Commented Jul 13, 2023 at 19:45
• @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. Commented Jul 14, 2023 at 7:13