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

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

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integral of sec^2(θ)e^{2tan(θ)}

\int sec^{2} (θ) e^{2 t a n (θ)} d θ

\frac{1}{2} e^{2 t a n (θ)} + C

Solution steps

\int sec^{2} (θ) e^{2 t a n (θ)} d θ

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= \int \frac{e ^{u}}{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 e^{u} d u

Use the common integral:

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

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

Substitute back

u = 2 tan (θ)

= \frac{1}{2} e^{2 t a n (θ)}

Add a constant to the solution

= \frac{1}{2} e^{2 t a n (θ)} + C

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