How do you find the antiderivative of $e^{2 x} \cdot {(tan (e^{2 x}))}^{2}$ $e^(2x)*( tan(e^(2x)) )^2$ ?

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How do you find the antiderivative of $e^{2 x} \cdot {(tan (e^{2 x}))}^{2}$ $e^(2x)*( tan(e^(2x)) )^2$ ?

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Answer 1 · 2016-10-22T20:28:52

$\frac{1}{2} tan (e^{2 x}) - \frac{1}{2} e^{2 x} + C$ $1/2tan(e^(2x))-1/2e^(2x)+C$

Explanation:

$I = \int e^{2 x} {tan}^{2} (e^{2 x}) d x$ $I=inte^(2x)tan^2(e^(2x))dx$

Let $u = e^{2 x}$ $u=e^(2x)$ , so that $d u = 2 e^{2 x} d x$ $du=2e^(2x)dx$ . Thus:

$I = \frac{1}{2} \int {tan}^{2} (e^{2 x}) (2 e^{2 x} d x)$ $I=1/2inttan^2(e^(2x))(2e^(2x)dx)$

$I = \frac{1}{2} \int {tan}^{2} (u) d u$ $I=1/2inttan^2(u)du$

We can integrate this using the identity ${tan}^{2} (u) + 1 = {sec}^{2} (u)$ $tan^2(u)+1=sec^2(u)$ :

$I = \frac{1}{2} \int ({sec}^{2} (u) - 1) d u$ $I=1/2int(sec^2(u)-1)du$

$I = \frac{1}{2} \int {sec}^{2} (u) d u - \frac{1}{2} \int d u$ $I=1/2intsec^2(u)du-1/2intdu$

These are both common integrals:

$I = \frac{1}{2} tan (u) - \frac{1}{2} u + C$ $I=1/2tan(u)-1/2u+C$

With $u = e^{2 x}$ $u=e^(2x)$ :

$I = \frac{1}{2} tan (e^{2 x}) - \frac{1}{2} e^{2 x} + C$ $I=1/2tan(e^(2x))-1/2e^(2x)+C$

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How do you find the antiderivative of $e^{2 x} \cdot {(tan (e^{2 x}))}^{2}$ $e^(2x)*( tan(e^(2x)) )^2$ ?

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How do you find the antiderivative of $e^{2 x} \cdot {(tan (e^{2 x}))}^{2}$ $e^(2x)*( tan(e^(2x)) )^2$ ?