What is the integral of e^{-x}tan(3+2e^{-x}) ?

The integral of e^{-x}tan(3+2e^{-x}) is 1/2 ln|cos(3+2e^{-x})|+C

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integral of e^{-x}tan(3+2e^{-x})

\int e^{- x} tan (3 + 2 e^{- x}) d x

\frac{1}{2} ln cos (3 + 2 e^{- x}) + C

Solution steps

\int e^{- x} tan (3 + 2 e^{- x}) d x

Apply u-substitution

: \int - \frac{tan ( u )}{2} d u

= \int - \frac{tan ( 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 tan (u) d u

Rewrite using trig identities

= - \frac{1}{2} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= - \frac{1}{2} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= - \frac{1}{2} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} (- ln ∣ v ∣)

Substitute back

= - \frac{1}{2} (- ln cos (3 + 2 e^{- x}))

Simplify

= \frac{1}{2} ln cos (3 + 2 e^{- x})

Add a constant to the solution

= \frac{1}{2} ln cos (3 + 2 e^{- x}) + C

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What is the integral of e^{-x}tan(3+2e^{-x}) ?
The integral of e^{-x}tan(3+2e^{-x}) is 1/2 ln|cos(3+2e^{-x})|+C