Evaluate
$$\int_0^a \frac{1}{{\cos x}\ {\cos (a-x)}}\mathrm{d}x$$
One solution begins by expanding the $\cos(a-x)$ term with the addition formula, dividing through by $\cos^2 x$ and substituting $u=\tan x$ leading to familiar log integral, real numbers only (given appropriate restrictions on $a$).
However my first attempt at this was to use the factor formula to get a sum of cosine terms in the denominator, and I thought this would work because one of the terms is independent of $x$, but it leads, via the Weierstrass ($\tan$ half-angle) substitution (unless I'm mistaken), to the following evaluation in $t$:
$$\frac{1}{i\sin a}\bigg[\tan^{-1}{(i\tan\frac{a}{2}}t)\bigg]_{\tan-\frac{a}{2}}^{\tan\frac{a}{2}}$$
I'm wondering whether this can be evaluated from here, and if so, whether the answer can be shown to be equal to the log function from the previous method.
(My complex analysis knowledge is very limited).
Thanks.
Best Answer
With real method\begin{align} &\int_0^a \frac{1}{{\cos x}\ {\cos (a-x)}}\ dx\\ =& \int_0^a \frac{2}{{\cos a}+{\cos (2x-a)}}dx =\int_{-a/2}^{a/2} \frac{2}{{\cos a}+{\cos 2x}}dx\\ =& \int_{-a/2}^{a/2} \frac{\sec^2 x}{{\cos^2 \frac a2}-\ {\sin^2 \frac a2 \ \tan^2 x}}dx\\ =& \ \frac2{\sin a}\tanh^{-1}\left(\tan\frac a2\tan x\right)_{-a/2}^{a/2}\\ =& \ \frac4{\sin a}\tanh^{-1}\tan^2\frac a2 = {2\csc a}\ln (\sec a)\\ \\ \end{align}
For the complex method, use $\tan^{-1}(ix) =i\tanh^{-1}x$ \begin{align}\frac{1}{i\sin a}\tan^{-1} \left(i\tan\frac{a}{2}t\right) =\frac{1}{\sin a}\tanh^{-1}\left(\tan\frac{a}{2}t\right)\\ \end{align}