What is the integral from 0 to pi/2 of cos^3(θ/2) ?

The integral from 0 to pi/2 of cos^3(θ/2) is 5/(3sqrt(2))

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integral from 0 to pi/2 of cos^3(θ/2)

\int_{0}^{\frac{π}{2}} cos^{3} (\frac{θ}{2}) d θ

\frac{5}{3 2}

Decimal Notation

1.17851 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{3} (\frac{θ}{2}) d θ

Apply u-substitution

: \int_{0}^{\frac{π}{4}} cos^{3} (u) \cdot 2 d u

= \int_{0}^{\frac{π}{4}} cos^{3} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int_{0}^{\frac{π}{4}} cos^{3} (u) d u

Rewrite using trig identities

= 2 \cdot \int_{0}^{\frac{π}{4}} (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int_{0}^{\frac{2}{2}} 1 - v^{2} d v

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{\frac{2}{2}} v^{2} d v = \frac{1}{6 2}

= 2 (\frac{1}{2} - \frac{1}{6 2})

Simplify

2 (\frac{1}{2} - \frac{1}{6 2}) : \frac{5}{3 2}

= \frac{5}{3 2}

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What is the integral from 0 to pi/2 of cos^3(θ/2) ?
The integral from 0 to pi/2 of cos^3(θ/2) is 5/(3sqrt(2))