Integral of exponential function

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$\int \left({\sec}^210x\;\tan10x\right)\;e^{{\tan}^210x} \; dx$

**Answer**

$\int \left({\sec}^210x\;\tan10x\right)\;e^{{\tan}^210x} \; dx$

Substitute $u= {\tan}^210x$. So, $du =20{\sec}^210x\tan10x\; dx$

And, $\frac{1}{20}\; du = {\sec}^2 10x\;\tan10x\; dx$

Then $\int \left({\sec}^210x\;\tan10x\right)\;e^{{\tan}^210x} \; dx$

$=\frac{1}{20}\int e^{u} \; du$

$=\frac{1}{20}e^u + C $

Substituting back $u= {\tan}^2 10x$

$=\frac{1}{20}e^{{\tan}^210x} + C$