Evaluate the integral $\int\frac{1}{\cos (x)+\cos (\alpha)}dx$

Question

Evaluate $$\int \frac{1}{\cos (x)+\cos (\alpha)}dx$$

My try: $$\int \frac{1}{\cos (x)+\cos (\alpha)}dx=\int \frac{(\cos x -\cos \alpha)}{\cos^2 (x)-\cos^2 (\alpha)}dx$$

Now $\displaystyle\int \frac{\cos x}{\cos^2 (x)-\cos^2 (\alpha)}dx$ can be calculated. What about the rest?

Answer 1

\begin{align} &\int \frac{1}{\cos x+\cos a}dx \\ =& \int\frac{1}{2\cos^2\frac x2-2\sin^2\frac a2}dx=\int\frac{\frac12\sec^2 \frac x2}{\cos^2\frac a2-\sin^2\frac a2\tan^2 \frac x2}dx\\ =&\int \frac{\ d\left(\tan \frac x2\right)}{\cos^2\frac a2-\sin^2\frac a2\tan^2 \frac x2} =\frac{2}{\sin\alpha} \tanh^{-1}\left(\tan\frac{a}2\tan\frac x2\right) \end{align}

Answer 2

Hint:

Let $z=e^{ix}$ and $a=e^{i\alpha}$.

$$I=\int\frac{2\,dz}{((z+z^{-1})-(a+a^{-1}))\,iz}=\int\frac{2\,dz}{(z^2+1-(a+a^{-1})z)\,i}=\int\frac{2\,dz}{(z-a)(z-a^{-1})\,i}.$$

Then decompose in simple fractions.

Answer 3

Applying the tangent half-angle substitution, $$x = 2 \arctan t, \qquad dx = \frac{2 \,dt}{1 + t^2},$$ transforms the integral to $$2 \int \frac{dt}{(\cos \alpha - 1) t^2 + \cos \alpha + 1} .$$

For $\alpha$ not an integer multiple of $\pi$ (so that $\cos \alpha \not\in \{\pm1\}$), substituting $$t = u \cot\frac\alpha2, \qquad dt = \cos \frac\alpha2\,du$$ transforms the integral to $$2 \csc \alpha \int \frac{du}{1 - u^2} = 2 \csc \alpha \operatorname{artanh} u + C = 2 \csc \alpha \operatorname{artanh} \left(\tan \frac\alpha2 \tan \frac{x}2\right) + C.$$

If $\cos \alpha = \pm 1$, we can exploit the Pythagorean identity: \begin{align}\int \frac{dx}{\cos x \pm 1} &= \int \frac{\cos \mp 1}{(\cos x \pm 1)(\cos x \mp 1)} dx \\ &= \int \frac{\pm 1 -\cos x}{\sin^2 x} dx \\ &= \int \left(\pm \csc^2 x - \csc x \cot x\right) dx \\ &= \mp \cot x + \csc x + C.\end{align} For $\pm = +$, we can rewrite this quantity as $\tan \frac x2 + C$, and for $\pm = -$, we can rewrite it as $\cot \frac x2 + C$.

Answer 4

Define:$$I_1=\int \frac{1}{\cos x+\cos \alpha}dx,~~~~~~~I_2=\int \frac{1}{\cos x-\cos \alpha}dx$$

We have

$$I_1+I_2=\int \frac{2\cos x}{\cos^2 x-\cos^2 \alpha}dx=\int \frac{2}{\sin^2 \alpha-\sin^2 x}d\sin x=\frac1{\sin\alpha}\cdot\ln\left|\frac{\sin x+\sin\alpha}{\sin x-\sin\alpha}\right|$$ and

$$\begin{align}I_1-I_2&=\int\frac{-2\cos\alpha}{\cos^2 x-\cos^2 \alpha\cdot\color{red}{(\sin^2 x+\cos^2 x)}}\cdot\color{blue}{\frac{\cos^2 x}{\cos^2 x}}dx\\ \\ &=\int\frac{-2\cos\alpha}{\sin^2\alpha-\cos^2\alpha\tan^2x}d\tan x=-\frac1{\sin\alpha}\cdot\ln\left|\frac{\tan x+\tan\alpha}{\tan x-\tan \alpha}\right|\end{align}$$

We are done.