Integral of exponential function

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$\int \left({\sec}^26x\;\tan6x\right)\;e^{{\tan}^26x} \; dx$

**Answer**

$\int \left({\sec}^26x\;\tan6x\right)\;e^{{\tan}^26x} \; dx$

Substitute $u= {\tan}^26x$. So, $du =12{\sec}^26x\tan6x\; dx$

And, $\frac{1}{12}\; du = {\sec}^2 6x\;\tan6x\; dx$

Then $\int \left({\sec}^26x\;\tan6x\right)\;e^{{\tan}^26x} \; dx$

$=\frac{1}{12}\int e^{u} \; du$

$=\frac{1}{12}e^u + C $

Substituting back $u= {\tan}^2 6x$

$=\frac{1}{12}e^{{\tan}^26x} + C$