What is the integral from 0 to pi/2 of xsin(8x) ?

The integral from 0 to pi/2 of xsin(8x) is -pi/(16)

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integral from 0 to pi/2 of xsin(8x)

\int_{0}^{\frac{π}{2}} x sin (8 x) d x

- \frac{π}{16}

Decimal Notation

- 0.19634 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x sin (8 x) d x

Apply Integration By Parts

: [\frac{1}{8} (- x cos (8 x) - 8 \cdot \int - \frac{1}{8} cos (8 x) d x)]_{0}^{\frac{π}{2}}

= [\frac{1}{8} (- x cos (8 x) - 8 \cdot \int - \frac{1}{8} cos (8 x) d x)]_{0}^{\frac{π}{2}}

\int - \frac{1}{8} cos (8 x) d x = - \frac{1}{64} sin (8 x)

= [\frac{1}{8} (- x cos (8 x) - 8 (- \frac{1}{64} sin (8 x)))]_{0}^{\frac{π}{2}}

Simplify

[\frac{1}{8} (- x cos (8 x) - 8 (- \frac{1}{64} sin (8 x)))]_{0}^{\frac{π}{2}} : [\frac{1}{8} (- x cos (8 x) + \frac{1}{8} sin (8 x))]_{0}^{\frac{π}{2}}

= [\frac{1}{8} (- x cos (8 x) + \frac{1}{8} sin (8 x))]_{0}^{\frac{π}{2}}

Compute the boundaries:

- \frac{π}{16}

= - \frac{π}{16}

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What is the integral from 0 to pi/2 of xsin(8x) ?
The integral from 0 to pi/2 of xsin(8x) is -pi/(16)