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

The integral from 0 to pi of sin^3(x/2) is 4/3

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integral from 0 to pi of sin^3(x/2)

\int_{0}^{π} sin^{3} (\frac{x}{2}) d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{π} sin^{3} (\frac{x}{2}) d x

Apply u-substitution

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

= \int_{0}^{\frac{π}{2}} sin^{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}{2}} sin^{3} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

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

\int_{0}^{1} v^{2} d v = \frac{1}{3}

= 2 (- (- 1 + \frac{1}{3}))

Simplify

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

= \frac{4}{3}

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What is the integral from 0 to pi of sin^3(x/2) ?
The integral from 0 to pi of sin^3(x/2) is 4/3