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

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Answer 1 · 2016-04-30T20:01:24

$lnabs(secx+tanx)+C$

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

$int1/cosxdx=intsecxdx$

This is an important trigonometric identity:

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

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

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

How do you find the integral of 1/cos x1/cos x?