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

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

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

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

Apply u-substitution

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

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

Substitute back

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

Simplify

2 \cdot \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

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