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

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

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

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

\frac{arctan ( \frac{1}{2} )}{2}

Decimal

0.43520 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int_{0}^{\frac{1}{2}} \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

arctan (\frac{1}{2})

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

Simplify

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

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