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

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

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

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

\frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{2}{u ^{2} + 9} d u

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

n = 6 \sum \infty \frac{5 ^{n}}{1 4 ^{n}}

\frac{d}{d x} (\frac{x}{x - 10})

ln (1 - z)

\int 3 x^{2} - 24 x + 48 d x

\int \frac{x ^{2} - 7 x - 12}{x ( x ^{2} + 3 x + 3 )} d x

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