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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 :

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

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 :

$$\int \frac{1}{1-u^2}\,\mathrm du = \int \frac{1/2}{1+u}\,\mathrm du + \int \frac{1/2}{1-u}\,\mathrm du = 1/2 (\int \frac{1}{1+u} \, du - \int \frac{1}{1-u} \, du) = \frac 1 2 \ln(\frac{1+u}{1-u}) = \ln(\frac{1+\sin(x)}{1-\sin(x)})^{1/2} = \ln(\frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}}) = \ln(\frac{\sqrt{1+\sin(x)}}{\sqrt{\cos(x)^2}}) = \ln(\frac{\sqrt{1+\sin(x)}}{\cos(x)})$$

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. $\endgroup$ – André Nicolas Jun 13 '13 at 7:43
  • $\begingroup$ @AndréNicolas Thanks for the hint. Could you see my edit please ? $\endgroup$ – user2336315 Jun 13 '13 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 Jun 13 '13 at 9:18
  • $\begingroup$ @Adriano Nice ! $\endgroup$ – user2336315 Jun 13 '13 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. $\endgroup$ – André Nicolas Jun 13 '13 at 14:18
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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 Oct 4 '18 at 12:45
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$$\int\frac{1}{1-u^2}du=\frac12\int(\frac{1}{1+u}+\frac{1}{1-u})du=\frac12(\ln(1+u)-\ln(1-u))+\color{red }{\ln c}=\ln(\color{red }{c}\sqrt{\frac{1+u}{1-u}})$$

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  • $\begingroup$ Thanks ! Upvoted ! Could you see my edit please ? $\endgroup$ – user2336315 Jun 13 '13 at 8:31
  • $\begingroup$ I don't get you. What do you mean by add its constant ? $\endgroup$ – user2336315 Jun 13 '13 at 8:40
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You can also remember that :$1/\cos(x)=\sec(x)$

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    $\begingroup$ What good will that do? ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 13 '15 at 20:20
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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). $\endgroup$ – MattAllegro Dec 13 '15 at 21:37
  • $\begingroup$ I didn't downvote by the way. $\endgroup$ – MattAllegro Dec 13 '15 at 21:38
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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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