Integral of exponential function

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$\int \left(\sec3x\;\tan3x\right)\;e^{\sec3x} \; dx$

**Answer**

$\int \left(\sec3x\;\tan3x\right)\;e^{\sec3x} \; dx$

Substitute $u= \sec3x$. So, $du =3\sec3x\;\tan3x\; dx$

And, $\frac{1}{3}\; du = \sec 3x\;\tan3x\; dx$

Then $\int \left(\sec3x\;\tan3x\right)\;e^{\sec3x} \; dx$

$=\frac{1}{3}\int e^{u} \; du$

$=\frac{1}{3}e^u + C $

Substituting back $u= \sec 3x$

$=\frac{1}{3}e^{\sec 3x} + C$