 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 $M^{\prime }=\sec xM$. The right-hand side after division is also $M^{\prime }$, soThe identity $(\sec x+\tan x)(\sec x-\tan x)=1$ shows $M$ never vanishes on the interval, givingAn integrating factor may have either fixed sign; if obtained by integrating $\sec x$, the real logarithm is $\ln |\sec x+\tan x|$.