Integral of exponential function

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$\int \left({\sec}^25x\;\tan5x\right)\;e^{{\tan}^25x} \; dx$

**Answer**

$\int \left({\sec}^25x\;\tan5x\right)\;e^{{\tan}^25x} \; dx$

Substitute $u= {\tan}^25x$. So, $du =10{\sec}^25x\tan5x\; dx$

And, $\frac{1}{10}\; du = {\sec}^2 5x\;\tan5x\; dx$

Then $\int \left({\sec}^25x\;\tan5x\right)\;e^{{\tan}^25x} \; dx$

$=\frac{1}{10}\int e^{u} \; du$

$=\frac{1}{10}e^u + C $

Substituting back $u= {\tan}^2 5x$

$=\frac{1}{10}e^{{\tan}^25x} + C$