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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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

= - \int_{0}^{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_{0}^{1} 1 d u + \int_{0}^{1} u^{2} d u)

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

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

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

Simplify

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

= \frac{2}{3}

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

\int_{0}^{8} 2 π x (x^{\frac{3}{2}}) d x

\int 8 x sec^{2} (x) d x

\int x sin (2 a x) d x

y = (1 + 6 x^{2}) (x - x^{2})

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