What is the integral of 1/(5+4cos(2x)) ?

The integral of 1/(5+4cos(2x)) is 1/3 arctan((tan(x))/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(5+4cos(2x))

\int \frac{1}{5 + 4 cos ( 2 x )} d x

\frac{1}{3} arctan (\frac{tan ( x )}{3}) + C

Solution steps

\int \frac{1}{5 + 4 cos ( 2 x )} d x

Apply u-substitution

: \int \frac{1}{2 ( 5 + 4 cos ( u ) )} d u

= \int \frac{1}{2 ( 5 + 4 cos ( u ) )} 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 \frac{1}{5 + 4 cos ( u )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 9} d v

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{tan ( \frac{2 x}{2} )}{3})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{tan ( \frac{2 x}{2} )}{3}) : \frac{1}{3} arctan (\frac{tan ( x )}{3})

= \frac{1}{3} arctan (\frac{tan ( x )}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{tan ( x )}{3}) + C

\int (5 x - 6) d x

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

\int 4 x^{2} (3 x + 1)^{2} d x

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

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

What is the integral of 1/(5+4cos(2x)) ?
The integral of 1/(5+4cos(2x)) is 1/3 arctan((tan(x))/3)+C