0
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by Carl Mummert, N. F. Taussig, Surb, user133281, Jack D'Aurizio Feb 23 '15 at 14:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ See this. $\endgroup$ – David Mitra Feb 23 '15 at 12:21
  • $\begingroup$ The goal of an indefinite integral is only to find the antiderivative. If you already know the antiderivative, try taking its derivative, and see how that simplifies down to $1/\cos(x)$. For simplicity, do this without the absolute value - usually that is found by doing a case analysis (or a substitution the leads to $\int 1/u\,du$). $\endgroup$ – Carl Mummert Feb 23 '15 at 12:21
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Liked this, thx) $\endgroup$ – dimaastronom Feb 23 '15 at 14:45
0
$\begingroup$

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

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Please format your answer using MathJax, a LaTeX-style system for typesetting math in answers. $\endgroup$ – Carl Mummert Feb 23 '15 at 12:22
  • 1
    $\begingroup$ Consider writing your answer with mathjax. A small guide about it can be found here $\endgroup$ – Bman72 Feb 23 '15 at 12:22
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thx, it was very helpful! $\endgroup$ – dimaastronom Feb 23 '15 at 12:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.