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

Question

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?

Answer 1

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?

Answer 2

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.

Answer 3

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.