How do you find the antiderivative of $\int tan x d x$ $int tanx dx$ ?

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Answer 1 · 2016-10-10T20:55:01

$ln | sec x | + C$ $lnabssecx+C" "$ or $- ln | cos x | + C$ $" "-lnabscosx+C$

Rewrite $tan x$ $tanx$ using its form in $sin x$ $sinx$ and $cos x$ $cosx$ :

$\int tan x d x = \int \frac{sin x}{cos x} d x$ $inttanxdx=intsinx/cosxdx$

Now, we can use the substitution $u = cos x$ $u=cosx$ . This implies that $d u = - sin x d x$ $du=-sinxdx$ .

$\int \frac{sin x}{cos x} = - \int \frac{- sin x}{cos x} d x = - \int \frac{d u}{u}$ $intsinx/cosx=-int(-sinx)/cosxdx=-int(du)/u$

This is a common and valuable integral to recognize:

$- \int \frac{d u}{u} = - ln | u | + C$ $-int(du)/u=-lnabs(u)+C$

Back-substituting with $u = cos x$ $u=cosx$ :

$- ln | u | + C = - ln | cos x | + C$ $-lnabsu+C=-lnabscosx+C$

Note that this can be rewritten using logarithm rules, with the negative on the outside being brought into the logarithm as a $- 1$ $-1$ power.

$- ln | cos x | + C = ln ∣ ∣ {(cos x)}^{- 1} ∣ ∣ + C = ln | sec x | + C$ $-lnabscosx+C=lnabs((cosx)^-1)+C=lnabssecx+C$

How do you find the antiderivative of ∫tanxdxint tanx dx?