Integral of exponential function
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Question

$\int \left({\sec}^212x\;\tan12x\right)\;e^{{\sec}^212x} \; dx$

$\int \left({\sec}^212x\;\tan12x\right)\;e^{{\sec}^212x} \; dx$

Substitute $u= {\sec}^212x$. So, $du =24{\sec}^212x\;\tan12x\; dx$

And, $\frac{1}{24}\; du = {\sec}^2 12x\;\tan12x\; dx$

Then $\int \left({\sec}^212x\;\tan12x\right)\;e^{{\sec}^212x} \; dx$

$=\frac{1}{24}\int e^{u} \; du$

$=\frac{1}{24}e^u + C$

Substituting back $u= {\sec}^2 12x$

$=\frac{1}{24}e^{{\sec}^2 12x} + C$