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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Simplify

\frac{tan ( e ^{u} )}{e ^{- u}} : e^{u} tan (e^{u})

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

Apply u-substitution

: \int tan (v) d v

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

Rewrite using trig identities

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

Apply u-substitution

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

= - \frac{1}{3} \cdot \int - \frac{1}{w} d w

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= - \frac{1}{3} (- ln ∣ w ∣)

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

\int \frac{t ^{2}}{- 7 + 6 t - t ^{2}} d t

\int \frac{cos ( x )}{7 sin ^{2} ( x ) + 8 sin ( x )} d x

\int \frac{1}{( e ^{x} + 1 ) ^{2} e ^{- x}} d x

\int \frac{x ^{2}}{2} \frac{1}{1 + x ^{2}} d x

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

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