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

The integral from 0 to pi/2 of 1/(3-2cos(x)) is 2/(sqrt(5))arctan(sqrt(5))

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

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

\frac{2}{5} arctan (5)

Decimal

1.02882 \dots

Solution steps

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

Apply u-substitution

= \int_{0}^{1} \frac{2}{5 u ^{2} + 1} 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}{5 u ^{2} + 1} d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{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}^{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}^{5}

Simplify

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

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

Compute the boundaries:

arctan (5)

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

What is the integral from 0 to pi/2 of 1/(3-2cos(x)) ?
The integral from 0 to pi/2 of 1/(3-2cos(x)) is 2/(sqrt(5))arctan(sqrt(5))