What is the integral from 0 to pi/2 of-1cos^5(x) ?

The integral from 0 to pi/2 of-1cos^5(x) is -8/15

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integral from 0 to pi/2 of-1cos^5(x)

\int_{0}^{\frac{π}{2}} - 1 cos^{5} (x) d x

- \frac{8}{15}

Decimal Notation

- 0.53333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} - 1 \cdot cos^{5} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

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

Apply u-substitution

: \int_{0}^{1} (1 - u^{2})^{2} d u

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

Expand

(1 - u^{2})^{2} : 1 - 2 u^{2} + u^{4}

= - \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})

Join

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

= - \frac{8}{15}

(5.1) (9.4)

n = 1 \sum \infty \frac{2}{5 ^{n}}

(9 x + 1) y^{2} \frac{d y}{d x} + 2 x^{2} + 3 y^{3} = 0

\frac{1}{x + 3}

y^{^{'}} = (x + y - 5)^{2}

What is the integral from 0 to pi/2 of-1cos^5(x) ?
The integral from 0 to pi/2 of-1cos^5(x) is -8/15