5
$\begingroup$

could you help me on this integral ?

$$\int \frac{1}{\cos(x)}\,\mathrm dx$$

Here's what I've started :

$$\int \frac{1}{\cos(x)}\,\mathrm dx = \int \frac{\cos(x)}{\cos(x)^2}\,\mathrm dx = \int \frac{\cos(x)}{1-\sin(x)^2}\,\mathrm dx$$

Now, I did : $u = \sin(x)$, so $\mathrm du = 1$.

Now I have :

$$\int \frac{\text{???}}{1-u^2}\,\mathrm du$$

But at this point, I think I did the most of the job but I'm stuck. Could you help me to solve this integral please (to the integration by substitution at the end) ?

Thanks

EDIT :

Now I follow the steps and I got :

$$\int \frac{1}{1-u^2}\,\mathrm du$$ Doing the partial fraction I got $A = 1/2$ and $B = 1/2$.

So basically I have :

\begin{align} & \int \frac{1}{1-u^2}\,\mathrm du = \int \frac{1/2}{1+u}\,\mathrm du + \int \frac{1/2}{1-u}\,\mathrm du \\[8pt] = {} & \frac 1 2 \left(\int \frac{1}{1+u} \, du - \int \frac{1}{1-u} \, du\right) \\[8pt] = {} & \frac 1 2 \ln\left(\frac{1+u}{1-u}\right) = \ln\left(\left(\frac{1+\sin(x)}{1-\sin(x)}\right)^{1/2}\right) \\[8pt] = {} & \ln \left(\frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}}\right) = \ln\left(\frac{\sqrt{1+\sin(x)}}{\sqrt{\cos(x)^2}}\right) \\[8pt] = {} & \ln\left(\frac{\sqrt{1+\sin(x)}}{\cos(x)}\right) \end{align}

Here's what my teacher got :

enter image description here

What's wrong with what I did ? Did I miss something ?

$\endgroup$
5
  • 7
    $\begingroup$ $du=\cos x\,dx$, which brings you to $\int\frac{du}{1-u^2}$. Now use partial fractions. $\endgroup$ Jun 13, 2013 at 7:43
  • $\begingroup$ @AndréNicolas Thanks for the hint. Could you see my edit please ? $\endgroup$ Jun 13, 2013 at 8:31
  • 1
    $\begingroup$ @user2336315 Note that since: $$\sec x + \tan x = \dfrac{1}{\cos x} + \dfrac{\sin x}{\cos x} = \dfrac{1+\sin x}{\cos x}$$ this is equivalent to my answer. Looking at your work, you should have done: $$ \sqrt{\dfrac{1+\sin x}{1 - \sin x}} = \sqrt{\dfrac{1+\sin x}{1 - \sin x} \cdot \dfrac{1+\sin x}{1 + \sin x}} = \sqrt{\dfrac{(1+\sin x)^2}{1 - \sin^2 x}} = \sqrt{\dfrac{(1+\sin x)^2}{\cos^2 x}} = \dfrac{1+\sin x}{\cos x} $$ $\endgroup$
    – Adriano
    Jun 13, 2013 at 9:18
  • $\begingroup$ @Adriano Nice ! $\endgroup$ Jun 13, 2013 at 9:38
  • $\begingroup$ Towards the end, when you "multiplied top and bottom by $1+\sin x$" you forgot to multiply the top. By the way, the conventional way of putting it is $\ln(|\sec x+\tan x|)$ and there is a quick excessively magic way to do it all in one line. $\endgroup$ Jun 13, 2013 at 14:18

7 Answers 7

11
$\begingroup$

Alternatively, observe that: $$ \int \dfrac{1}{\cos x} dx = \int \sec x~dx = \int \sec x \left(\dfrac{\sec x + \tan x}{\sec x + \tan x}\right) dx = \int\dfrac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx $$ Now let $u=\sec x + \tan x$ so that $du = (\sec x \tan x + \sec^2x)~dx$. Then we obtain: $$ \int\dfrac{(\sec x \tan x + \sec^2 x)~dx}{\sec x + \tan x} = \int \dfrac{du}{u}=\ln|u|+C= \boxed{\ln|\sec x + \tan x|+C} $$

$\endgroup$
1
  • 6
    $\begingroup$ whilst this is a well-known method, it is difficult to do if you do not know it $\endgroup$
    – Henry Lee
    Oct 4, 2018 at 12:45
4
$\begingroup$

\begin{align} & \int\frac{1}{1-u^2}\,du=\frac12\int\left(\frac{1}{1+u}+\frac{1}{1-u}\right) \,du \\[8pt] = {} & \frac12(\ln(1+u)-\ln(1-u))+\color{red }{\ln c} = \ln\left(\color{red }{c}\sqrt{\frac{1+u}{1-u}} \, \right) \end{align}

$\endgroup$
2
  • $\begingroup$ Thanks ! Upvoted ! Could you see my edit please ? $\endgroup$ Jun 13, 2013 at 8:31
  • $\begingroup$ I don't get you. What do you mean by add its constant ? $\endgroup$ Jun 13, 2013 at 8:40
1
$\begingroup$

$$ \frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}} = \frac{\sqrt{1+\sin(x)}\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}\sqrt{1+\sin(x)}} = \frac{1+\sin(x)}{\sqrt{\cos(x)^2}} =\text{etc.} $$

$\endgroup$
1
$\begingroup$

You can also remember that :$1/\cos(x)=\sec(x)$

$\endgroup$
1
  • 6
    $\begingroup$ What good will that do? ${}\qquad{}$ $\endgroup$ Dec 13, 2015 at 20:20
0
$\begingroup$

I know this is an old post but I spotted the mistake in your solution after the edit.

You wrote 1-sin(x) = cos(x)^2 (which is wrong for obvious reasons)

$\endgroup$
2
  • $\begingroup$ This would probably be better as a comment (I think). $\endgroup$ Dec 13, 2015 at 21:37
  • $\begingroup$ I didn't downvote by the way. $\endgroup$ Dec 13, 2015 at 21:38
0
$\begingroup$

Here is another try: $$I=\int\sec(x)dx=\int\frac{1}{\cos(x)}dx$$ now $u=\cos(x)$ so $dx=\frac{du}{-\sin(x)}$ then: $$I=-\int\frac{1}{\sqrt{1-u^2}u}du$$ then we can use $v=1-u^2$ so $du=\frac{dv}{-2u}$ and the integral becomes: $$I=\frac{1}{2}\int\frac{1}{v(1-v)}dv$$ and by using partial fraction decomposition we can obtain: $$\frac{1}{v(1-v)}=\frac{1}{v}+\frac{1}{1-v}$$ and so our integral becomes: $$I=\frac{1}{2}\int\frac{1}{v}+\frac{1}{1-v}dv=\frac{1}{2}\ln|\frac{v}{1-v}|+C=\frac{1}{2}\ln|\frac{1-u^2}{u^2}|+C=\frac{1}{2}\ln|\frac{1-\cos^2(x)}{\cos^2(x)}|+C$$ although this looks wrong to me

$\endgroup$
0
$\begingroup$

Notice how$$\left[\sec x\right]’_x=\sec x\tan x$$$$\left[\tan x\right]’_x=\sec^2 x$$A factor of $\sec x$ shows up in both results of the derivative! Adding the two equations together gives$$\left[\sec x\right]’_x+\left[\tan x\right]’_x=\sec x(\tan x+\sec x)$$Divide both sides by $\sec x+\tan x$ and integrate with respect to $x$. The right-hand side becomes the integral you’re looking$$\begin{align*}\int\mathrm dx\,\sec x & =\int\mathrm dx\,\frac {\left[\sec x\right]’_x+\left[\tan x\right]’_x}{\tan x+\sec x}\\ & =\int\frac {\mathrm du}u\\ & =\log(\sec x+\tan x)+C\end{align*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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