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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{2}{v ^{2} + 3} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{v ^{2} + 3} d v

Apply Integral Substitution

= 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

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}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{cos ^{3} ( x )}{sin ^{3} ( x ) + 8} d x

\int \frac{x ^{\frac{2}{3}}}{( x ^{\frac{1}{3}} + 1 ) ^{2}} d x

\int (3 x - 6)^{2} d x

\int \frac{x - 2}{x ^{2} - 2 x + 2} d x

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