What is the integral from 0 to pi/2 of-9cos^5(x) ?

The integral from 0 to pi/2 of-9cos^5(x) is -24/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of-9cos^5(x)

\int_{0}^{\frac{π}{2}} - 9 cos^{5} (x) d x

- \frac{24}{5}

Decimal

- 4.8

Solution steps

\int_{0}^{\frac{π}{2}} - 9 cos^{5} (x) d x

Take the constant out:

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

= - 9 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (x) d x

Rewrite using trig identities

= - 9 \cdot \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

= - 9 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

Expand

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

= - 9 \cdot \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

= - 9 (\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}

= - 9 (1 - \frac{2}{3} + \frac{1}{5})

Simplify

- 9 (1 - \frac{2}{3} + \frac{1}{5}) : - \frac{24}{5}

= - \frac{24}{5}

\int \frac{2 x - 5}{x ^{2} - 5 x - 1} d x

\frac{\partial}{\partial x} (sin (x y) + x cos (y))

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{9}}})

y = 2 x + 5, y = 14 - x^{2}, (- 1, 2)

y = x^{\frac{1}{2}}, y = x^{2}

What is the integral from 0 to pi/2 of-9cos^5(x) ?
The integral from 0 to pi/2 of-9cos^5(x) is -24/5