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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

  1. $$\sin x \tan x + \cos x = \sin x \left(\frac {\sin x}{\cos x}\right) + \cos x = \frac {\sin^2x}{\cos x}+\cos x$$
  2. $$\frac {\sin^2x}{\cos x}+\cos x \left(\frac {\cos x}{\cos x}\right)= \frac {\cos^2x + \sin^2x}{\cos x} = \frac 1{\cos x} = \sec x$$
  3. $$\frac {\sin^2x}{\cos x}+\cos x \,=\, \frac 12 \left(\frac {1-\cos {2x}}{\cos x} \right) + \cos x \,=\, \frac 1{2\cos x} - \frac {\cos {2x}\sec x}2 + \cos x$$

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?

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  • $\begingroup$ Do you consider something like $\sqrt{\cos(2x)}$ valid or it has to be a rational function? $\endgroup$ – kingW3 Feb 12 '17 at 19:03
  • $\begingroup$ @kingW3 Only considering rational functions at this point, I should add this to my question, thank you. $\endgroup$ – user409521 Feb 12 '17 at 19:16
  • $\begingroup$ Also, $(\sec x)'= \sec x \tan x$, so there is that. $\endgroup$ – Alex Ortiz Feb 12 '17 at 19:27
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You can always add $\sin^2 x+\cos ^2x-1$ and the value wouldn't change. Also considering the formula $$\sin(n\theta)=\sum_{\text{k is odd}}(-1)^{(k-1)/2}{n\choose k}\cos^{n-k}\theta \sin^k\theta $$ Taking $\theta=\frac{x}{n}$ one can get an infinite number of representations for $\sin x$,which is present in the solution(there is a formula for $\cos$ also).Presumably using substitution(s) like $\sin(x/n)=t$ one could get similar forms of the integral for every integer $n$.

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Short way $$\int \sec x\tan x\, dx=-\int \frac { d\left( \cos { x } \right) }{ \cos ^{ 2 }{ x } } =\frac { 1 }{ \cos { x } } +C$$

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Just to add to your list, try using the well known substitution $t=\tan\frac 12x$

The integral becomes $$\int \frac{4t}{(1-t^2)^2}dt=\frac{2}{1-t^2}+c$$

This is the same as your answers modulo a constant

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Any result when brought to its simplest trigonometric form should be $ \sec x +c.$

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