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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 ?

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    $\begingroup$ $du=\cos x\,dx$, which brings you to $\int\frac{du}{1-u^2}$. Now use partial fractions. $\endgroup$ Commented Jun 13, 2013 at 7:43
  • $\begingroup$ @AndréNicolas Thanks for the hint. Could you see my edit please ? $\endgroup$ Commented Jun 13, 2013 at 8:31
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    $\begingroup$ @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
    Commented Jun 13, 2013 at 9:18
  • $\begingroup$ @Adriano Nice ! $\endgroup$ Commented 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. $\endgroup$ Commented Jun 13, 2013 at 14:18

10 Answers 10

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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: $$ \int\dfrac{(\sec x \tan x + \sec^2 x)~dx}{\sec x + \tan x} = \int \dfrac{du}{u}=\ln|u|+C= \boxed{\ln|\sec x + \tan x|+C} $$

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

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

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  • $\begingroup$ This would probably be better as a comment (I think). $\endgroup$ Commented Dec 13, 2015 at 21:37
  • $\begingroup$ I didn't downvote by the way. $\endgroup$ Commented Dec 13, 2015 at 21:38
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You can also remember that :$1/\cos(x)=\sec(x)$

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    $\begingroup$ What good will that do? ${}\qquad{}$ $\endgroup$ Commented Dec 13, 2015 at 20:20
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$$ \frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}} = \frac{\sqrt{1+\sin(x)}\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}\sqrt{1+\sin(x)}} = \frac{1+\sin(x)}{\sqrt{\cos(x)^2}} =\text{etc.} $$

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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

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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$$\begin{align*}\int\mathrm dx\,\sec x & =\int\mathrm dx\,\frac {\left[\sec x\right]’_x+\left[\tan x\right]’_x}{\tan x+\sec x}\\ & =\int\frac {\mathrm du}u\\ & =\log(\sec x+\tan x)+C\end{align*}$$

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$$ \begin{aligned} \int \sec d x &=\int \frac{\cos x}{\cos ^{2} x} d x \\ &=\int \frac{1}{(1+\sin x)(1-\sin x)} d(\sin x) \\ &=\frac{1}{2} \int \left(\frac{1}{1-\sin x}+\frac{1}{1+\sin x}\right) d (\sin x) \\ &=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x} \right| +C \\ &=\frac{1}{2} \ln \left|\frac{(1+\sin x)^{2}}{\cos ^{2} x}\right| +C\\ &=\ln \left|\frac{1+\sin x}{\cos x}\right| +C\\&=\ln |\sec x+\tan x|+C \end{aligned} $$

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$\displaystyle \frac{1}{\cos{x}}=\frac{\cos{x}}{\cos^{2}{x}}=\frac{\cos{x}}{1-\sin^{2}{x}}$

$\displaystyle \sin{x}=u\Rightarrow du=\cos{x} dx$

$\displaystyle \frac{dx}{\cos{x}}=\frac{\ du}{1-\ u^{2}}=\frac{1}{2}(\frac{1}{u-1}-\frac{1}{u+1})$

$\displaystyle \int\frac{dx}{\cos{x}}=\ln\sqrt{ \frac{{\sin{x}-1}}{{\sin{x}+1}}}+c$

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Actually, the standard approach (Euler formula) when integrating trigonometric functions can be used, but it will take a different form, i.e., \begin{align} \int{\frac{1}{\cos x}dx}&=\int\frac{2e^{ix}}{e^{2ix}+1}dx\\ &=-2i\tan^{-1}(e^{ix})+C\\ &=\ln\left|\frac{1-ie^{ix}}{1+ie^{ix}}\right|+C \end{align}

which is equivalent to $\ln\sqrt{\left|\frac{\sin x+1}{\sin x-1}\right|}+C$ since \begin{align} \sqrt{\left|\frac{\sin x+1}{\sin x-1}\right|}= \sqrt{\left|\frac{e^{ix}-e^{-ix}+2i}{e^{ix}-e^{-ix}-2i}\right|}= \sqrt{\left|\frac{e^{2ix}-1+2ie^{ix}}{e^{2ix}-1-2ie^{ix}}\right|}= \sqrt{\left(\frac{e^{ix}+i}{e^{ix}-i}\right)^2}=\left|\frac{1-ie^{ix}}{1+ie^{ix}}\right| \end{align}

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