What is the integral from 0 to 2 of 1/(3+2cos(x)) ?

The integral from 0 to 2 of 1/(3+2cos(x)) is (2arctan(\frac{tan(1))/(sqrt(5)))}{sqrt(5)}

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integral from 0 to 2 of 1/(3+2cos(x))

\int_{0}^{2} \frac{1}{3 + 2 cos ( x )} d x

\frac{2 arctan ( \frac{t a n ( 1 )}{5} )}{5}

Decimal Notation

0.54414 \dots

Solution steps

\int_{0}^{2} \frac{1}{3 + 2 cos ( x )} d x

Apply u-substitution

: \int_{0}^{t a n (1)} \frac{2}{u ^{2} + 5} d u

= \int_{0}^{t a n (1)} \frac{2}{u ^{2} + 5} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{t a n (1)} \frac{1}{u ^{2} + 5} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{\frac{t a n ( 1 )}{5}} \frac{1}{5 ( 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}{5} \cdot \int_{0}^{\frac{t a n ( 1 )}{5}} \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}{5} [arctan (v)]_{0}^{\frac{t a n ( 1 )}{5}}

Simplify

2 \cdot \frac{1}{5} [arctan (v)]_{0}^{\frac{t a n ( 1 )}{5}} : \frac{2}{5} [arctan (v)]_{0}^{\frac{1}{5} t a n (1)}

= \frac{2}{5} [arctan (v)]_{0}^{\frac{1}{5} t a n (1)}

Compute the boundaries:

arctan (\frac{1}{5} tan (1))

= \frac{2}{5} arctan (\frac{1}{5} tan (1))

Simplify

= \frac{2 arctan ( \frac{t a n ( 1 )}{5} )}{5}

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What is the integral from 0 to 2 of 1/(3+2cos(x)) ?
The integral from 0 to 2 of 1/(3+2cos(x)) is (2arctan(\frac{tan(1))/(sqrt(5)))}{sqrt(5)}