What is the integral of 1/(2+cos(x)) ?

The integral of 1/(2+cos(x)) is 2/(sqrt(3))arctan(1/(sqrt(3))tan(x/2))+C

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integral of 1/(2+cos(x))

\int \frac{1}{2 + cos ( x )} d x

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

Solution steps

\int \frac{1}{2 + cos ( x )} d x

Apply u-substitution

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

= \int \frac{2}{u ^{2} + 3} 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} + 3} 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 = arctan (v)

= 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 2 \cdot \frac{1}{3} arctan (\frac{tan ( \frac{x}{2} )}{3})

Simplify

2 \cdot \frac{1}{3} arctan (\frac{tan ( \frac{x}{2} )}{3}) : \frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2}))

= \frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2}))

Add a constant to the solution

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

y^{^{'}} = x y^{\frac{1}{3}}

\frac{\partial}{\partial t} (t)

\frac{5 ^{x}}{x ^{5}}

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{7}}})

f (x) = x + 2 \cdot x + 2

What is the integral of 1/(2+cos(x)) ?
The integral of 1/(2+cos(x)) is 2/(sqrt(3))arctan(1/(sqrt(3))tan(x/2))+C