What is the integral from 0 to pi/2 of 1/(3+5cos(x)) ?

The integral from 0 to pi/2 of 1/(3+5cos(x)) is (ln(3))/4

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integral from 0 to pi/2 of 1/(3+5cos(x))

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 5 cos ( x )} d x

\frac{ln ( 3 )}{4}

Decimal Notation

0.27465 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 5 cos ( x )} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{- u ^{2} + 4} d u

= \int_{0}^{1} \frac{1}{- u ^{2} + 4} 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 = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Simplify

\frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}} : \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}}

= \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

= \frac{1}{2} \cdot \frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

Simplify

= \frac{ln ( 3 )}{4}

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What is the integral from 0 to pi/2 of 1/(3+5cos(x)) ?
The integral from 0 to pi/2 of 1/(3+5cos(x)) is (ln(3))/4