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

The integral from 0 to pi/2 of xsin(3x) is -1/9

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integral from 0 to pi/2 of xsin(3x)

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

- \frac{1}{9}

Decimal

- 0.11111 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{3} (- x cos (3 x) - 3 \cdot \int - \frac{1}{3} cos (3 x) d x)]_{0}^{\frac{π}{2}}

= [\frac{1}{3} (- x cos (3 x) - 3 \cdot \int - \frac{1}{3} cos (3 x) d x)]_{0}^{\frac{π}{2}}

\int - \frac{1}{3} cos (3 x) d x = - \frac{1}{9} sin (3 x)

= [\frac{1}{3} (- x cos (3 x) - 3 (- \frac{1}{9} sin (3 x)))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{3} (- x cos (3 x) - 3 (- \frac{1}{9} sin (3 x)))]_{0}^{\frac{π}{2}} : [\frac{1}{3} (- x cos (3 x) + \frac{1}{3} sin (3 x))]_{0}^{\frac{π}{2}}

= [\frac{1}{3} (- x cos (3 x) + \frac{1}{3} sin (3 x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

- \frac{1}{9}

= - \frac{1}{9}

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What is the integral from 0 to pi/2 of xsin(3x) ?
The integral from 0 to pi/2 of xsin(3x) is -1/9