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

The integral from 0 to pi of xcos(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 of xcos(3x)

\int_{0}^{π} x cos (3 x) d x

- \frac{2}{9}

Solution steps

\int_{0}^{π} x cos (3 x) d x

Apply u-substitution

: \int_{0}^{3 π} \frac{u cos ( u )}{9} d u

= \int_{0}^{3 π} \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_{0}^{3 π} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{3 π}

= \frac{1}{9} [u sin (u) - \int sin (u) d u]_{0}^{3 π}

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

= \frac{1}{9} [u sin (u) - (- cos (u))]_{0}^{3 π}

Simplify

= \frac{1}{9} [u sin (u) + cos (u)]_{0}^{3 π}

Compute the boundaries:

- 2

= \frac{1}{9} (- 2)

Simplify

= - \frac{2}{9}

t^{2} y^{^{''}} + 9 t y^{^{'}} + 7 y = 0

\int \frac{x}{x ^{2} - 3 x + 2} d x

\int \frac{y ^{2} + 2 y + 1}{( y ^{2} + 1 ) ^{2}} d y

4 y^{^{''}} + 28 y^{^{'}} + 49 y = 0

f (t) = e^{t + 9}

What is the integral from 0 to pi of xcos(3x) ?
The integral from 0 to pi of xcos(3x) is -2/9