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

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

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integral from 0 to pi/2 of sin^5(θ)

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

\frac{8}{15}

Decimal

0.53333 \dots

Solution steps

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

Rewrite using trig identities

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

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}

\int_{0}^{2 π} (cos (x))^{6} d x

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\int_{8}^{11} x d x

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

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