Question

How do you find the integral of `1/cos x`?

Answer 1

`lnabs(secx+tanx)+C`

Explanation:

Note that `1/cosx=secx`. Thus, we see that

`int1/cosxdx=intsecxdx`

This is an important trigonometric identity:

`intsecxdx=lnabs(secx+tanx)+C`

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If you want to know how to find `intsecxdx`, the process is unintuitive, but I've outlined it below:

`intsecxdx=int(secx(secx+tanx))/(secx+tanx)dx=int(secxtanx+sec^2x)/(secx+tanx)dx`

Now, use substitution:

`u=secx+tanx" "=>" "du=(secxtanx+sec^2x)dx`

Note that these are the fraction's numerator and denominator.

`int(secxtanx+sec^2x)/(secx+tanx)dx=int(du)/u=lnabsu+C`

Since `u=secx+tanx`, we obtain

`intsecxdx=lnabs(secx+tanx)+C`