If we use the definition of tanx and write the integral as
tanxdx=sinxcosxdx
we see that the numerator sinx is the derivative of the denominator (apart from the minus sign). Hence, the substitution u=cosx will work,
sinxcosxdx=u=cosxdu=cosxdx=−sinxdx=−udu=−lnu+C=−lncosx+C
NOTE: −lncosx+C is only a primitive function in intervals in which cosx=0.
tanxdx=sinxcosxdx sinxcosxdx=
u=cosxdu=
cosx
dx=−sinxdx
=−udu=−ln
u+C=−lncosx+C cosx+C
=0