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 $$\int (1+1+1)\,dx = \int 3\, dx = 3x+C$$