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

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

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

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

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

Solution steps

\int \frac{1}{4 - 5 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

Factor

9 u^{2} - 1 : - (- 9 u^{2} + 1)

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

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 (- \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 2 (- \frac{1}{3} (\frac{ln 3 tan ( \frac{x}{2} ) + 1}{2} - \frac{ln 3 tan ( \frac{x}{2} ) - 1}{2}))

Simplify

2 (- \frac{1}{3} (\frac{ln 3 tan ( \frac{x}{2} ) + 1}{2} - \frac{ln 3 tan ( \frac{x}{2} ) - 1}{2})) : - \frac{1}{3} (ln 3 tan (\frac{x}{2}) + 1 - ln 3 tan (\frac{x}{2}) - 1)

= - \frac{1}{3} (ln 3 tan (\frac{x}{2}) + 1 - ln 3 tan (\frac{x}{2}) - 1)

Add a constant to the solution

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

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