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

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

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

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

\frac{π}{3 3}

Decimal

0.60459 \dots

Solution steps

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

Apply u-substitution

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

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

Complete the square

u^{2} + 1 + u : (u + \frac{1}{2})^{2} + \frac{3}{4}

= \int_{0}^{1} \frac{1}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Apply u-substitution

: \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4}{4 v ^{2} + 3} d v

= \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4}{4 v ^{2} + 3} d v

Take the constant out:

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

= 4 \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 v ^{2} + 3} d v

Apply Integral Substitution

= 4 \cdot \int_{\frac{1}{3}}^{3} \frac{1}{2 3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 4 \cdot \frac{1}{2 3} \cdot \int_{\frac{1}{3}}^{3} \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 4 \cdot \frac{1}{2 3} [arctan (w)]_{\frac{1}{3}}^{3}

Simplify

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

= \frac{2}{3} [arctan (w)]_{\frac{1}{3}}^{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+sin(x)) ?
The integral from 0 to pi/2 of 1/(2+sin(x)) is pi/(3sqrt(3))