What is the integral from 0 to pi of sin^5(x/2) ?

The integral from 0 to pi of sin^5(x/2) is 16/15

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integral from 0 to pi of sin^5(x/2)

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

\frac{16}{15}

Decimal

1.06666 \dots

Solution steps

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin^{5} (u) \cdot 2 d u

= \int_{0}^{\frac{π}{2}} sin^{5} (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^{5} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

= 2 \cdot \int_{1}^{0} - 1 + 2 v^{2} - v^{4} d v

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

= 2 (- \int_{0}^{1} - 1 + 2 v^{2} - v^{4} 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} 2 v^{2} d v - \int_{0}^{1} v^{4} d v))

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

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

\int_{0}^{1} v^{4} d v = \frac{1}{5}

= 2 (- (- 1 + \frac{2}{3} - \frac{1}{5}))

Simplify

2 (- (- 1 + \frac{2}{3} - \frac{1}{5})) : \frac{16}{15}

= \frac{16}{15}

t \to π lim (sin (\frac{t}{6}))

\frac{d y}{d x} = 2 y 16 - x^{2}, y (0) = 1

\frac{d}{d x} (ln (16 - x^{2}))

\int 2 (2 + x^{2})^{\frac{3}{2}} d x

\int x^{2} (x^{3} + 6)^{6} d x

What is the integral from 0 to pi of sin^5(x/2) ?
The integral from 0 to pi of sin^5(x/2) is 16/15