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

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

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integral from 0 to-pi/3 of xcos(3x)

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

- \frac{2}{9}

Decimal Notation

- 0.22222 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

2

= - \frac{1}{9} \cdot 2

Simplify

= - \frac{2}{9}

\int_{1}^{4} \frac{5}{x} d x

\int_{1}^{e} 5 t^{2} ln (t) d t

\frac{d x}{d t} = x - x^{2}, x (0) = 2

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} - 6 x = 4 e^{- 2 t}

\int - 36 cos (x) d x

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