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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Simplify

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

Add a constant to the solution

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

y^{^{''}} - y^{^{'}} + y = cos (2 t) + sin (2 t)

\frac{1}{s ( s ^{2} + 6 π s + π ^{2} )}

\int \frac{3 tan ( x ) - 4 cos ^{2} ( x )}{cos ( x )} d x

5 x^{2} (3 x^{2} - 1)^{2}

\int ((x - 1)^{5} + 3 (x - 1)^{2} + 5) d x

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