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1.30 Continuing Wright’s integral (1599)

Note that $\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}=\sec x+\tan x.$ Indeed,

\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}=\frac{\cos\frac{x}{2}+\sin\frac{x}% {2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}=\frac{(\cos\frac{x}{2}+\sin\frac{x}{2})^% {2}}{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}

=\frac{\cos^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}+\sin^{2}\frac{x}{2}% }{\cos x}=\frac{1+\sin{x}}{\cos{x}}=\sec x+\tan x,

using the double angle formulae.

Thus we can also say: $\int\sec x\,dx=\log|\sec x+\tan x|+C$ .