What is the integral of sec^2(θ)e^{4tan(θ)} ?

The integral of sec^2(θ)e^{4tan(θ)} is 1/4 e^{4tan(θ)}+C

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integral of sec^2(θ)e^{4tan(θ)}

\int sec^{2} (θ) e^{4 t a n (θ)} d θ

\frac{1}{4} e^{4 t a n (θ)} + C

Solution steps

\int sec^{2} (θ) e^{4 t a n (θ)} d θ

Apply u-substitution

: \int e^{4 u} d u

= \int e^{4 u} d u

Apply u-substitution

: \int e^{v} \frac{1}{4} d v

= \int e^{v} \frac{1}{4} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{4} e^{v}

Substitute back

= \frac{1}{4} e^{4 t a n (θ)}

Add a constant to the solution

= \frac{1}{4} e^{4 t a n (θ)} + C

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What is the integral of sec^2(θ)e^{4tan(θ)} ?
The integral of sec^2(θ)e^{4tan(θ)} is 1/4 e^{4tan(θ)}+C