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

The integral from 0 to pi/2 of 1/(2+cos(x)) is pi/(3sqrt(3))

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

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

\frac{π}{3 3}

Decimal Notation

0.60459 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{2}{u ^{2} + 3} d u

= \int_{0}^{1} \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_{0}^{1} \frac{1}{u ^{2} + 3} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{\frac{1}{3}} \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_{0}^{\frac{1}{3}} \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)]_{0}^{\frac{1}{3}}

Simplify

2 \cdot \frac{1}{3} [arctan (v)]_{0}^{\frac{1}{3}} : \frac{2}{3} [arctan (v)]_{0}^{\frac{1}{3}}

= \frac{2}{3} [arctan (v)]_{0}^{\frac{1}{3}}

Compute the boundaries:

\frac{π}{6}

= \frac{2}{3} \cdot \frac{π}{6}

Simplify

= \frac{π}{3 3}

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