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

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

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integral from 0 to pi/6 of xcos(3x)

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

\frac{π - 2}{18}

Decimal

0.06342 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

\frac{π}{2} - 1

= \frac{1}{9} (\frac{π}{2} - 1)

Simplify

= \frac{π - 2}{18}

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