Integral of Tangent function

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$\int \tan\left(2x-4\right)\; dx$

**Answer**

$\int \tan\left(2x-4\right)\; dx$

Substitute $u=2x-4$. So, $du =2\; dx$

Or, $\frac{1}{2}\; du = dx$

And, $\int \tan\left(2x-4\right)\; dx = \frac{1}{2}\int \tan u \;du$

$=-\frac{1}{2}\ln \left|\cos u\right| + C$

Substituting back $u=2x-4$

$=-\frac{1}{2}\ln \left| \cos\left(2x-4\right)\right|+C$