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

The integral from 0 to pi/6 of cos^3(3x) is 2/9

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 cos^3(3x)

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

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

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

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

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

= \frac{1}{3} (1 - \frac{1}{3})

Simplify

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

= \frac{2}{9}

\int_{- 6}^{6} 2 + 2 e^{2 t} d t

\int_{1}^{7} x ln (x) d x

\int_{0}^{2 s i n (θ)} r d r

\int_{- 3}^{- 2} (2 x - 1) d x

\int_{1}^{5} (- x + 10) d x

What is the integral from 0 to pi/6 of cos^3(3x) ?
The integral from 0 to pi/6 of cos^3(3x) is 2/9