= Solution
Divide by $\cos x$ on an interval where it is nonzero. For the resulting <first-order linear differential equation>, use the <integrating factor> $M=\sec x+\tan x$, since $M'=\sec x\,M$. The right-hand side after division is also $M'$, so
$$
(My)'=MM',\qquad My=\frac12M^2+C.
$$
The identity $(\sec x+\tan x)(\sec x-\tan x)=1$ shows $M$ never vanishes on the interval, giving
$$
\boxed{y=\frac12(\sec x+\tan x)+C(\sec x-\tan x).}
$$
An <integrating factor> may have either fixed sign; if obtained by integrating $\sec x$, the real <logarithm> is $\ln|\sec x+\tan x|$.
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