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 Notation

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 e^{2 x} x + \frac{3}{2} e^{2 x} d x

n = 1 \sum \infty \frac{3 n}{5 n + 1}

2 x, 2 x - 4, 0, 4

\int - 2 t^{2} d t

\frac{\partial}{\partial x} (cos (4 x + 5 y) \cdot 5)

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