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

The integral from 0 to-pi/6 of xcos(3x) is -(2-pi)/(18)

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 xcos(3x)

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

- \frac{2 - π}{18}

Decimal Notation

0.06342 \dots

Solution steps

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

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

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

Apply u-substitution

: \int_{- \frac{π}{2}}^{0} \frac{u cos ( u )}{9} d u

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

Take the constant out:

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

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

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{- \frac{π}{2}}^{0}

= - \frac{1}{9} [u sin (u) - \int sin (u) d u]_{- \frac{π}{2}}^{0}

\int sin (u) d u = - cos (u)

= - \frac{1}{9} [u sin (u) - (- cos (u))]_{- \frac{π}{2}}^{0}

Simplify

= - \frac{1}{9} [u sin (u) + cos (u)]_{- \frac{π}{2}}^{0}

Compute the boundaries:

1 - \frac{π}{2}

= - \frac{1}{9} (1 - \frac{π}{2})

Simplify

= - \frac{2 - π}{18}

y = 8 x + 7

t \to 1 lim (t + 8)

\frac{\partial}{\partial x} (4 \cdot e^{x} - 2 \cdot x \cdot e^{y})

\frac{d}{d x} ((4 x - 1)^{\frac{1}{3}})

\frac{x ^{4} + 6 x - 1}{7 x ^{2} - 6 x + 1}

What is the integral from 0 to-pi/6 of xcos(3x) ?
The integral from 0 to-pi/6 of xcos(3x) is -(2-pi)/(18)