What is the integral of e^{-4x}tan(e^{-4x}) ?

The integral of e^{-4x}tan(e^{-4x}) is 1/4 ln|cos(e^{-4x})|+C

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integral of e^{-4x}tan(e^{-4x})

\int e^{- 4 x} tan (e^{- 4 x}) d x

\frac{1}{4} ln cos (e^{- 4 x}) + C

Solution steps

\int e^{- 4 x} tan (e^{- 4 x}) d x

Apply u-substitution

= \int - \frac{tan ( u )}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int tan (u) d u

Rewrite using trig identities

= - \frac{1}{4} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

= - \frac{1}{4} \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}{4} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{4} (- ln ∣ v ∣)

Substitute back

= - \frac{1}{4} (- ln cos (e^{- 4 x}))

Simplify

= \frac{1}{4} ln cos (e^{- 4 x})

Add a constant to the solution

= \frac{1}{4} ln cos (e^{- 4 x}) + C

What is the integral of e^{-4x}tan(e^{-4x}) ?
The integral of e^{-4x}tan(e^{-4x}) is 1/4 ln|cos(e^{-4x})|+C