The easiest way to integrate $\tan(x)$ is to recall that $$ \tan(x) = \frac{\sin(x)}{\cos(x)}\,, $$ so $$ ∫ \tan(x)\,dx = ∫ \frac{\sin(x)}{\cos(x)}\,dx\,. $$ Do you see the necessary substitution? By choosing $u = \cos(x)$, that is, “$du = -\sin(x)\,dx$” (in quotation marks because this expression does not make sense mathematically, but it does work formally), we get $$ ∫ \frac{\sin(x)}{\cos(x)}\,dx = ∫ -\frac{1}{u}\,du = -\log(u)+c = -\log(\cos(x))+c $$

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