I am always astounded by number of equivalent ways to express trigonometric expression and I wonder how many ways there are to solve integrals such as $$\int \sec x \tan x \, dx$$ I use the equivalence that feels intuitive to me, $$\int \sec x \tan x \, dx \, = \, \int \left(\frac 1{\cos x}\right)\left(\frac {\sin x}{\cos x}\right)\,dx$$ $$=\, \int \frac{\sin x}{\cos^2x}\,dx \,=\, \int \sec^2x \sin x \, dx$$ then use integration by parts, $$\int \sec^2x \sin x \, dx \,=\, \sin x \tan x - \int \cos x \tan x \, dx$$ Again I use what feels intuitive and substitute $\int \cos x \tan x \, dx$ for $$\int \cos x\left( \frac {\sin x}{\cos x}\right) \, dx = \int \sin x \, dx \,=\, -\cos x$$ Therefore, $$\int \sec x \tan x \, dx = \int \sec^2x \sin x \, dx \,=\, \sin x \tan x + \cos x +C$$ From there I try to find equivalent forms (I will omit writing the arbitrary constant every time), so I get for example
- $$\sin x \tan x + \cos x = \sin x \left(\frac {\sin x}{\cos x}\right) + \cos x = \frac {\sin^2x}{\cos x}+\cos x$$
- $$\frac {\sin^2x}{\cos x}+\cos x \left(\frac {\cos x}{\cos x}\right)= \frac {\cos^2x + \sin^2x}{\cos x} = \frac 1{\cos x} = \sec x$$
- $$\frac {\sin^2x}{\cos x}+\cos x \,=\, \frac 12 \left(\frac {1-\cos {2x}}{\cos x} \right) + \cos x \,=\, \frac 1{2\cos x} - \frac {\cos {2x}\sec x}2 + \cos x$$
It seems to me that I could always multiply a factor of $ 1= \left(\frac {trig-expression}{trig-expression}\right)$ to any result and find an equivalent expression, of higher or lower complexity than the original expression.
Hence, only considering rational functions, I wonder if that substitution "game" can be played $ad\;infinitum$, or is there a strictly finite number of equivalent forms or ways to express results involving trig-expressions? In the case that number is finite, can it be computed?