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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

Rewrite using trig identities

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

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}

\int_{0}^{2} (x^{2} - 1) - (1 - x^{2}) d x

\int_{0}^{5} t e^{- 0.4 t} d t

\int_{0}^{2} - x^{2} + 2 x + 1 d x

\int_{0}^{1} x + x d x

\int_{x}^{π - x} sin (x + y) d y

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