What is the integral from 0 to pi/6 of 4cos^5(3x) ?

The integral from 0 to pi/6 of 4cos^5(3x) is 32/45

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/6 of 4cos^5(3x)

\int_{0}^{\frac{π}{6}} 4 cos^{5} (3 x) d x

\frac{32}{45}

Decimal Notation

0.71111 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 4 cos^{5} (3 x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{6}} cos^{5} (3 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos^{5} (u) \frac{1}{3} d u

= 4 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (u) \frac{1}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} cos^{5} (u) d u

Rewrite using trig identities

= 4 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} (1 - sin^{2} (u))^{2} cos (u) d u

Apply u-substitution

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

= 4 \cdot \frac{1}{3} \cdot \int_{0}^{1} (1 - v^{2})^{2} d v

Expand

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

= 4 \cdot \frac{1}{3} \cdot \int_{0}^{1} 1 - 2 v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{3} (\int_{0}^{1} 1 d v - \int_{0}^{1} 2 v^{2} d v + \int_{0}^{1} v^{4} d v)

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} 2 v^{2} d v = \frac{2}{3}

\int_{0}^{1} v^{4} d v = \frac{1}{5}

= 4 \cdot \frac{1}{3} (1 - \frac{2}{3} + \frac{1}{5})

Simplify

4 \cdot \frac{1}{3} (1 - \frac{2}{3} + \frac{1}{5}) : \frac{32}{45}

= \frac{32}{45}

\int \frac{2}{x} + 3 sin (x) - 4 e^{x} d x

\frac{\partial}{\partial x} (6 z e^{x yz})

y^{^{'}} = 20 - 4 y

e^{- 43 t} \cdot e^{72 t} cos (77 t)

\int 5 sin (\frac{x}{4}) d x

What is the integral from 0 to pi/6 of 4cos^5(3x) ?
The integral from 0 to pi/6 of 4cos^5(3x) is 32/45