# Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$

Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$

My Attempt:

Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$ The given integral can be transformed as $$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \int\dfrac{\tan(x-3)}{\cos(x-3)}\,\textrm dx$$

The right most integral can be calculated easily by writing $$\tan(x-3)$$ as $$\frac{\sin(x-3)}{\cos(x-3)}$$ and then by a substituiton $$\cos(x-3)$$ as $$t$$. But I have no clue for the left most integral. How to evaluate that?

• Proper notation is not $\displaystyle \int\frac{1}{\mathrm{cos(x-1)cos(x-2)cos(x-3)}} \, dx,$ but $$\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)} \, dx,$$ coded as  \int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)} \, dx . I edited the question accordingly. When  \cos is used, then the horizontal spacing depends on the context, so that more space is to the right of $\cos$ in $\cos x$ than in $\cos(x). \qquad$ May 23, 2021 at 15:28

Hint:

I found the expression $$=\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}-\dfrac{\sin(x-2)}{\cos(x-2)\cos(x-3)}$$

Now,

$$\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}$$

$$=\dfrac{\sin(x-1)}{\sin2}\cdot\dfrac{\sin(x-1-(x-3))}{\cos(x-1)\cos(x-3)}$$

$$=\dfrac{1-\cos^2(x-1)}{\sin2\cos(x-1)}-\dfrac{\sin(x-1)\sin(x-3)}{\sin2\cos(x-3)}$$

The first part can be managed easily.

For the second part, $$\dfrac{\sin(x-1)\sin(x-3)}{\cos(x-3)}=\dfrac{\sin(x-3+2)\sin(x-3)}{\cos(x-3)}=\dfrac{\cos2\sin^2(x-3)}{\cos(x-3)}-\sin2\sin(x-3)$$

Can you take it home from here?

Note $$\cos(x-n) = \cos x\cos n + \sin x \sin n$$ and rewrite the integrand as \begin{align} \frac{1}{{\cos(x-1)\cos(x-2)\cos(x-3)}} ={}\frac{\csc1\csc2\csc3\sec^3 x}{(\tan x+\cot 1) (\tan x+\cot 2) (\tan x+\cot 3)} \end{align} Substitute $$t=\tan x$$ and perform the partial fractionization

\begin{align} &\int \frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,dx \\ ={}& \frac{1}{\sin 2}\int\frac{\sqrt{1+t^2}}{t+\cot1}\,dt -\frac{1}{\sin1}\int\frac{\sqrt{1+t^2}}{t+\cot2}\,dt + \frac{\sin 3}{\sin1\sin2}\int\frac{\sqrt{1+t^2}}{t+\cot3} \, dt\\ \end{align} The three integrals are of the same form and can be readily carried out.

• Someone with a 56.7k reputation should have some LaTeX literacy. Look: \begin{align} & a+\cdots + z \\ = & a+\cdots+z \end{align} Do you not see the conspicuous lack of proper spacing after the "equals" sign? Now look at this: \begin{align} & a+\cdots + z \\ = {} & a+\cdots+z \end{align} Notice the difference in the code. (And of course, the visible difference in the result.) Similarly: \begin{align} & 5+3 \\ {} \\ & 5+ \\ {} \\ & {+3} \end{align} The space to the left and right of the "plus" sign is different in the three lines above. The software is deliberately$\,$ May 23, 2021 at 15:48
• $\ldots\,$ designed that way for a reason. The short version of the reason is that when Donald Knuth originally created TeX, he understood what he was doing. May 23, 2021 at 15:51
• @MichaelHardy Relax, Michael. This is Mathematics SE, not Tex SE. As long as the post is clear and readable it's not a huge deal. Edit the post if you want, but don't give a giant lecture in the comments that is irrelevant to the actual content of the answer. May 23, 2021 at 16:00
• @K.defaoite : But people who do things that way here also do them in other contexts. I see it all the time. A professor writes $a|b$ and $a\not| b$ instead of $a\mid b$ and $a\nmid b,$ and all sorts of other stuff like that. Some sort of literacy campaign might not be a bad idea. May 23, 2021 at 16:10

Little bit of trigonometry \begin{align} \frac{\tan(x-1)}{\cos(x-2)} &= \frac{\tan(x-1)}{\cos((x-1) - 1)} \\ &= \frac{\tan(x-1)}{\cos(x-1)\cos 1 + \sin(x-1)\sin 1} \\ &= \frac{\tan(x-1)\sec(x-1)}{\cos 1 + \tan(x-1)\sin 1} \\ &= \frac{d\left(\sec(x-1)\right)}{\cos1 + \sin 1 \sqrt{\sec^2(x-1) - 1}} \\ &= \frac{du}{\cos 1 + \sin 1 \sqrt{u^2-1}}\end{align} So, we are to calculate in a more general case \begin{align} \int \frac{du}{a+b\sqrt{u^2-1}} \end{align} You can find a solution for this here and it gives $$\frac{1}{2b}\ln\frac{|\sqrt{u^2-1}-1|}{|\sqrt{u^2-1}+1|} + \frac{a}{b\sqrt{a^2+b^2}}\left(\text{artanh}\frac{bu}{\sqrt{b^2+a^2}} - \text{artanh}\frac{\sqrt{b^2+a^2}\sqrt{u^2-1}}{au}\right) + C$$

Now substitute back $$\sqrt{u^2 - 1} = \tan(x-1)$$ and $$a = \cos 1$$, $$b = \sin 1$$ to get something prettier.

• I'm a highschool student and we aren't inteoduced to hyperbolic functions yet. Is it not possible without hyperbolic functions? May 23, 2021 at 15:58