What is the integral from 0 to (3pi)/2 of cos^3(33x) ?

The integral from 0 to (3pi)/2 of cos^3(33x) is -2/99

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to (3pi)/2 of cos^3(33x)

\int_{0}^{\frac{3 π}{2}} cos^{3} (33 x) d x

- \frac{2}{99}

Solution steps

\int_{0}^{\frac{3 π}{2}} cos^{3} (33 x) d x

Apply u-substitution

: \int_{0}^{\frac{99 π}{2}} cos^{3} (u) \frac{1}{33} d u

= \int_{0}^{\frac{99 π}{2}} cos^{3} (u) \frac{1}{33} d u

Take the constant out:

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

= \frac{1}{33} \cdot \int_{0}^{\frac{99 π}{2}} cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{33} \cdot \int_{0}^{\frac{99 π}{2}} (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= \frac{1}{33} \cdot \int_{0}^{- 1} 1 - v^{2} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{33} (- \int_{- 1}^{0} 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}{33} (- (\int_{- 1}^{0} 1 d v - \int_{- 1}^{0} v^{2} d v))

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

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

= \frac{1}{33} (- (1 - \frac{1}{3}))

Simplify

\frac{1}{33} (- (1 - \frac{1}{3})) : - \frac{2}{99}

= - \frac{2}{99}

\int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 x ^{2}} d x

\frac{\partial}{\partial y} (\frac{x ^{2}}{y + 2})

\frac{d}{d x} (tan (2 x^{3}))

\int \frac{x}{x ^{4} - 16 x ^{2} + 39} d x

y = x + x^{2} + x^{3}, x = 1

What is the integral from 0 to (3pi)/2 of cos^3(33x) ?
The integral from 0 to (3pi)/2 of cos^3(33x) is -2/99