What is the integral from 0 to pi of sin^3(t) ?

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

integral from 0 to pi of sin^3(t)

\int_{0}^{π} sin^{3} (t) d t

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{π} sin^{3} (t) d t

Rewrite using trig identities

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

Apply u-substitution

= \int_{1}^{- 1} - 1 + u^{2} d u

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

= - \int_{- 1}^{1} - 1 + u^{2} d u

Apply the Sum Rule:

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

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

\int_{- 1}^{1} 1 d u = 2

\int_{- 1}^{1} u^{2} d u = \frac{2}{3}

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

Simplify

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

= \frac{4}{3}

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