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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{tan ( u )}{3} d u

= \int - \frac{tan ( u )}{3} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int tan (u) d u

Rewrite using trig identities

= - \frac{1}{3} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= - \frac{1}{3} \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}{3} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{3} (- ln ∣ v ∣)

Substitute back

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

Simplify

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

Add a constant to the solution

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

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f (t) = t^{2} + 1

\frac{\partial}{\partial t} (s^{2} t)

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