Integral of square of Secant function

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$\int {\sec}^2\left(14x-8\right)\; dx$

**Answer**

$\int {\sec}^2\left(14x-8\right)\; dx$

Substitute $u=14x-8$. So, $du =14\; dx$

Or, $\frac{1}{14}\; du = dx$

And, $\int {\sec}^2\left(14x-8\right)\; dx = \frac{1}{14}\int {\sec u}^2 \;du$

$=\frac{1}{14}\tan u + C$

Substituting back $u=14x-8$

$=\frac{1}{14}\tan\left(14x-8\right)+C$