Find integral $\int\frac{1}{\cos x}dx$

Question

Need help with this integral $$\int\frac{1}{\cos x}dx$$ I know that the answer is $$\ln|\operatorname{tg} x+\sec x|$$ I tried transforming 1 into $\cos^2x + \sin^2x$ but it led to nothing. Need to solve it using simplest way without new variables and differential transformations.

Answer 1

This is my way of finding this integral: $$\begin{align}\int \frac{1}{\cos{x}}dx =\int \frac{\cos{x}}{\cos^2{x}} dx= \int \frac{\cos{x}}{1-\sin^2{x}} dx\\ \text{substitution } \Big|\begin{array}{cc}\sin{x}=u \\ \cos{x}dx=du\end{array} \Big| \\= \int \frac{1}{1-u^2}du = \tanh^{-1}u +C= \tanh^{-1}(\sin{x})+C\end{align}.$$

Answer 2

1/COSX=secx, multiply and divide it by secx+tanx you get int(secx^2+secxtanx)/(secx+tanx) dx which gives =log(tanx+secx)

Answer 3

Note that: $$ \dfrac{d}{dx} \tan x= \sec^2 x $$ and $$ \dfrac{d}{dx} \sec x= \sec x \tan x $$ so that: $$ d(\tan x +\sec x)= \dfrac{\tan x +\sec x}{\cos x} dx $$ and: $$ \dfrac{d(\tan x +\sec x)}{\tan x +\sec x}=\dfrac{dx}{\cos x} $$ so, integrating you have the result.