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

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

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

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

\frac{2}{3}

Solution steps

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

Rewrite using trig identities

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

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}

n = 2 \sum \infty \frac{n}{3 ^{n + 1}}

ln (\frac{8 - x}{8 + x})

72 x - \frac{4 x ^{3}}{3}

\frac{\partial}{\partial y} (\frac{y}{1 + x ^{2}})

\int \frac{1}{x + 2} d x

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