What is the integral of 1/(3+5cos(x)) ?

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

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integral of 1/(3+5cos(x))

\int \frac{1}{3 + 5 cos ( x )} d x

\frac{1}{4} (ln \frac{tan ( \frac{x}{2} )}{2} + 1 - ln \frac{tan ( \frac{x}{2} )}{2} - 1) + C

Solution steps

\int \frac{1}{3 + 5 cos ( x )} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 4} d u

= \int \frac{1}{- u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \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}

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

Substitute back

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

Simplify

\frac{1}{2} \frac{ln \frac{t a n ( \frac{x}{2} )}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} )}{2} - 1}{2} : \frac{1}{4} (ln \frac{tan ( \frac{x}{2} )}{2} + 1 - ln \frac{tan ( \frac{x}{2} )}{2} - 1)

= \frac{1}{4} (ln \frac{tan ( \frac{x}{2} )}{2} + 1 - ln \frac{tan ( \frac{x}{2} )}{2} - 1)

Add a constant to the solution

= \frac{1}{4} (ln \frac{tan ( \frac{x}{2} )}{2} + 1 - ln \frac{tan ( \frac{x}{2} )}{2} - 1) + C

\frac{d}{d x} (x^{3} e^{5 x})

\int (x + \frac{1}{6 x}) d x

\int_{0}^{1} x (x^{2} - x) d x

\int \frac{x}{y} d y

n = 0 \sum \infty \frac{- 1 ^{n}}{n !}

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