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

The integral from 0 to pi/3 of sin^3(x) is 5/24

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

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

\frac{5}{24}

Decimal Notation

0.20833 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{3}} (1 - cos^{2} (x)) sin (x) d x

Apply u-substitution

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

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

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

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

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

\int_{\frac{1}{2}}^{1} u^{2} d u = \frac{7}{24}

= - (- \frac{1}{2} + \frac{7}{24})

Simplify

- (- \frac{1}{2} + \frac{7}{24}) : \frac{5}{24}

= \frac{5}{24}

\int (1 - r) r d r

\frac{\partial}{\partial y} (- 2 x)

\int e^{- x} \cdot x d x

\int 6^{s i n (x)} cos (x) d x

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 5 x}{x + 1})

What is the integral from 0 to pi/3 of sin^3(x) ?
The integral from 0 to pi/3 of sin^3(x) is 5/24