Integral of exponential function

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$\int \left(\sec6x\;\tan6x\right)\;e^{\sec6x} \; dx$

**Answer**

$\int \left(\sec6x\;\tan6x\right)\;e^{\sec6x} \; dx$

Substitute $u= \sec6x$. So, $du =6\sec6x\;\tan6x\; dx$

And, $\frac{1}{6}\; du = \sec 6x\;\tan6x\; dx$

Then $\int \left(\sec6x\;\tan6x\right)\;e^{\sec6x} \; dx$

$=\frac{1}{6}\int e^{u} \; du$

$=\frac{1}{6}e^u + C $

Substituting back $u= \sec 6x$

$=\frac{1}{6}e^{\sec 6x} + C$