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Integrating the secant requires a bit of manipulation.
Multiply secx by secx+tanxsecx+tanx, which is really the same as multiplying by 1. Thus, we have
∫(secx(secx+tanx)secx+tanx)dx
∫sec2x+secxtanxsecx+tanxdx
Now, make the following substitution:
u=secx+tanx
du=(secxtanx+sec2x)dx=(sec2x+secxtanx)dx
We see that du appears in the numerator of the integral, so we may apply the substitution:
∫duu=ln|u|+C
Rewrite in terms of x to get
∫secxdx=ln|secx+tanx|+C
This is an integral worth memorizing.
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