Integral of exponential function

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$\int \left({\sec}^215x\;\tan15x\right)\;e^{{\tan}^215x} \; dx$

**Answer**

$\int \left({\sec}^215x\;\tan15x\right)\;e^{{\tan}^215x} \; dx$

Substitute $u= {\tan}^215x$. So, $du =30{\sec}^215x\tan15x\; dx$

And, $\frac{1}{30}\; du = {\sec}^2 15x\;\tan15x\; dx$

Then $\int \left({\sec}^215x\;\tan15x\right)\;e^{{\tan}^215x} \; dx$

$=\frac{1}{30}\int e^{u} \; du$

$=\frac{1}{30}e^u + C $

Substituting back $u= {\tan}^2 15x$

$=\frac{1}{30}e^{{\tan}^215x} + C$