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

The integral of 1/(1+4cos(x)) is 1/(sqrt(15))(ln|sqrt(3/5)tan(x/2)+1|-ln|sqrt(3/5)tan(x/2)-1|)+C

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

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

\frac{1}{15} (ln \frac{3}{5} tan (\frac{x}{2}) + 1 - ln \frac{3}{5} tan (\frac{x}{2}) - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{- 3 u ^{2} + 5} d u

= \int \frac{2}{- 3 u ^{2} + 5} 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}{- 3 u ^{2} + 5} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{15 ( - 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}{15} \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 \cdot \frac{1}{15} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 2 \cdot \frac{1}{15} \frac{ln \frac{3}{5} tan ( \frac{x}{2} ) + 1}{2} - \frac{ln \frac{3}{5} tan ( \frac{x}{2} ) - 1}{2}

Simplify

2 \cdot \frac{1}{15} \frac{ln \frac{3}{5} tan ( \frac{x}{2} ) + 1}{2} - \frac{ln \frac{3}{5} tan ( \frac{x}{2} ) - 1}{2} : \frac{1}{15} (ln \frac{3}{5} tan (\frac{x}{2}) + 1 - ln \frac{3}{5} tan (\frac{x}{2}) - 1)

= \frac{1}{15} (ln \frac{3}{5} tan (\frac{x}{2}) + 1 - ln \frac{3}{5} tan (\frac{x}{2}) - 1)

Add a constant to the solution

= \frac{1}{15} (ln \frac{3}{5} tan (\frac{x}{2}) + 1 - ln \frac{3}{5} tan (\frac{x}{2}) - 1) + C

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What is the integral of 1/(1+4cos(x)) ?
The integral of 1/(1+4cos(x)) is 1/(sqrt(15))(ln|sqrt(3/5)tan(x/2)+1|-ln|sqrt(3/5)tan(x/2)-1|)+C