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

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

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integral of sec^2(x)e^{1+tan(x)}

\int sec^{2} (x) e^{1 + t a n (x)} d x

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

Solution steps

\int sec^{2} (x) e^{1 + t a n (x)} d x

sec^{2} (x) e^{1 + t a n (x)} = e^{1} sec^{2} (x) e^{t a n (x)}

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

Take the constant out:

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

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

Simplify

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

Apply u-substitution

: \int e^{u} d u

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

Use the common integral:

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

= e e^{u}

Substitute back

u = tan (x)

= e e^{t a n (x)}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e e^{t a n (x)} = e^{1 + t a n (x)}

= e^{1 + t a n (x)}

Add a constant to the solution

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

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