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

The integral of 1/(5-4cos(x)) is 2/3 arctan(3tan(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 (3 tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{2}{9 u ^{2} + 1} 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}{9 u ^{2} + 1} 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 (3 tan (\frac{x}{2}))

Simplify

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

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

Add a constant to the solution

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

t y^{^{'}} = t + y

\frac{d}{d x} (y^{2} - cos (x))

\frac{( s + 5 )}{s ( s + 4 ) ^{2}}

\int \frac{cos ( x )}{5 + sin ^{2} ( x )} d x

\frac{d y}{d x} = 6 e^{x - y}

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