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

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

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

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

\frac{8}{15}

Decimal Notation

0.53333 \dots

Solution steps

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

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= - \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}

\int (10 x)^{n} d x

\int_{2}^{5} 2 x^{3} d x

\frac{d y}{d x} = - y^{3}

y = x^{2} - 8 x + 9, (1, 2)

\frac{d}{d x} (u e^{- x})

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