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

The integral of 1/(2-cos(x)) is 2/(sqrt(3))arctan(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 (3 tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{3 u ^{2} + 1} d u

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

e^{x} (1 + x)^{- 1}

\frac{\partial}{\partial x} (2 x + 2 y)

f (x) = \frac{1}{x - 3}

\int \frac{( x + 6 )}{x ^{2} + 9} d x

\frac{s + 1}{s ^{2} - 3 s + 2}

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