What is the integral of e^{-7x}tan(e^{-7x}) ?

The integral of e^{-7x}tan(e^{-7x}) is 1/7 ln|cos(e^{-7x})|+C

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integral of e^{-7x}tan(e^{-7x})

\int e^{- 7 x} tan (e^{- 7 x}) d x

\frac{1}{7} ln cos (e^{- 7 x}) + C

Solution steps

\int e^{- 7 x} tan (e^{- 7 x}) d x

Apply u-substitution

: \int - \frac{tan ( u )}{7} d u

= \int - \frac{tan ( u )}{7} d u

Take the constant out:

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

= - \frac{1}{7} \cdot \int tan (u) d u

Rewrite using trig identities

= - \frac{1}{7} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

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

= - \frac{1}{7} \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}{7} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{7} (- ln ∣ v ∣)

Substitute back

= - \frac{1}{7} (- ln cos (e^{- 7 x}))

Simplify

= \frac{1}{7} ln cos (e^{- 7 x})

Add a constant to the solution

= \frac{1}{7} ln cos (e^{- 7 x}) + C

x y^{^{'}} + 2 y ln (x) = 3

\int_{1}^{3} \frac{x ^{2} + 8}{4 x - x ^{2}} d x

y^{^{'}} = x y

y = ln (x + x^{2} - 1)

x \to 0 + lim (ln (\frac{1}{x}))

What is the integral of e^{-7x}tan(e^{-7x}) ?
The integral of e^{-7x}tan(e^{-7x}) is 1/7 ln|cos(e^{-7x})|+C