What is the integral of sec^2(2x)e^{tan(2x)} ?

The integral of sec^2(2x)e^{tan(2x)} is 1/2 e^{tan(2x)}+C

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integral of sec^2(2x)e^{tan(2x)}

\int sec^{2} (2 x) e^{t a n (2 x)} d x

\frac{1}{2} e^{t a n (2 x)} + C

Solution steps

\int sec^{2} (2 x) e^{t a n (2 x)} d x

Apply u-substitution

: \int sec^{2} (u) e^{t a n (u)} \frac{1}{2} d u

= \int sec^{2} (u) e^{t a n (u)} \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sec^{2} (u) e^{t a n (u)} d u

Apply u-substitution

: \int e^{v} d v

= \frac{1}{2} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{2} e^{v}

Substitute back

= \frac{1}{2} e^{t a n (2 x)}

Add a constant to the solution

= \frac{1}{2} e^{t a n (2 x)} + C

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What is the integral of sec^2(2x)e^{tan(2x)} ?
The integral of sec^2(2x)e^{tan(2x)} is 1/2 e^{tan(2x)}+C