What is the integral of 2tan(4x) ?

The integral of 2tan(4x) is -1/2 ln|cos(4x)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2tan(4x)

\int 2 tan (4 x) d x

- \frac{1}{2} ln ∣ cos (4 x) ∣ + C

Solution steps

\int 2 tan (4 x) d x

Take the constant out:

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

= 2 \cdot \int tan (4 x) d x

Rewrite using trig identities

= 2 \cdot \int \frac{sin ( 4 x )}{cos ( 4 x )} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= 2 \cdot \int - \frac{1}{4 u} d u

Take the constant out:

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

= 2 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = cos (4 x)

= 2 (- \frac{1}{4} ln ∣ cos (4 x) ∣)

Simplify

2 (- \frac{1}{4} ln ∣ cos (4 x) ∣) : - \frac{1}{2} ln ∣ cos (4 x) ∣

= - \frac{1}{2} ln ∣ cos (4 x) ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ cos (4 x) ∣ + C

\frac{x f ( x )}{5}

\int \frac{( 5 x + 3 )}{( x ^{2} - 9 )} d x

\frac{d y}{d x} = \frac{( - y ( x + y + 1 ) )}{x + 2 y}

\int_{- 1}^{1} (1 + 1 - x^{2}) d x

t^{2} y^{^{''}} + 19 t y^{^{'}} + 162 y = 0

What is the integral of 2tan(4x) ?
The integral of 2tan(4x) is -1/2 ln|cos(4x)|+C