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

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

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

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

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

Solution steps

\int \frac{1}{4 + 5 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 = \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 \frac{t a n ( \frac{x}{2} )}{3} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} )}{3} - 1}{2}

Simplify

2 \cdot \frac{1}{3} \frac{ln \frac{t a n ( \frac{x}{2} )}{3} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} )}{3} - 1}{2} : \frac{1}{3} (ln \frac{tan ( \frac{x}{2} )}{3} + 1 - ln \frac{tan ( \frac{x}{2} )}{3} - 1)

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

Add a constant to the solution

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

\int cos (x) e^{a x} d x

t (ln (2 t))^{2}

3 x (a^{2} - x^{2})

lo g_{8} (x^{2} - 3 x)

\int (ln (x + 1)) d x

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