Integral of Tangent function

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$\int \tan\left(5x+9\right)\; dx$

**Answer**

$\int \tan\left(5x+9\right)\; dx$

Substitute $u=5x+9$. So, $du =5\; dx$

Or, $\frac{1}{5}\; du = dx$

And, $\int \tan\left(5x+9\right)\; dx = \frac{1}{5}\int \tan u \;du$

$=-\frac{1}{5}\ln \left|\cos u\right| + C$

Substituting back $u=5x+9$

$=-\frac{1}{5}\ln \left| \cos\left(5x+9\right)\right|+C$