Integral of exponential function

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$\int \left({\sec}^27x\;\tan7x\right)\;e^{{\sec}^27x} \; dx$

**Answer**

$\int \left({\sec}^27x\;\tan7x\right)\;e^{{\sec}^27x} \; dx$

Substitute $u= {\sec}^27x$. So, $du =14{\sec}^27x\;\tan7x\; dx$

And, $\frac{1}{14}\; du = {\sec}^2 7x\;\tan7x\; dx$

Then $\int \left({\sec}^27x\;\tan7x\right)\;e^{{\sec}^27x} \; dx$

$=\frac{1}{14}\int e^{u} \; du$

$=\frac{1}{14}e^u + C $

Substituting back $u= {\sec}^2 7x$

$=\frac{1}{14}e^{{\sec}^2 7x} + C$