What is the integral from 0 to pi/2 of (sin(x))^5 ?

The integral from 0 to pi/2 of (sin(x))^5 is 8/15

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integral from 0 to pi/2 of (sin(x))^5

\int_{0}^{\frac{π}{2}} (sin (x))^{5} d x

\frac{8}{15}

Decimal Notation

0.53333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} (sin (x))^{5} d x

Simplify

= \int_{0}^{\frac{π}{2}} sin^{5} (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - 1 + 2 u^{2} - u^{4} d u

= \int_{1}^{0} - 1 + 2 u^{2} - u^{4} d u

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

= - \int_{0}^{1} - 1 + 2 u^{2} - u^{4} 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} 2 u^{2} d u - \int_{0}^{1} u^{4} d u)

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

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

\int_{0}^{1} u^{4} d u = \frac{1}{5}

= - (- 1 + \frac{2}{3} - \frac{1}{5})

Simplify

- (- 1 + \frac{2}{3} - \frac{1}{5}) : \frac{8}{15}

= \frac{8}{15}

\frac{d y}{d x} - y = \frac{11}{8} e^{- \frac{x}{3}}

x \to - 4 lim (\frac{6}{x + 5})

\int - x^{2} + 16 d x

\frac{\partial}{\partial x} (ln (4 x + 7 y + 3 z))

\frac{d}{d x} (2 cos (x) + 2 sin (x))

What is the integral from 0 to pi/2 of (sin(x))^5 ?
The integral from 0 to pi/2 of (sin(x))^5 is 8/15