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

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

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integral from 0 to 2pi of xsin(x)cos(x)

\int_{0}^{2 π} x sin (x) cos (x) d x

- \frac{π}{2}

Decimal

- 1.57079 \dots

Solution steps

\int_{0}^{2 π} x sin (x) cos (x) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} x sin (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{2 π} x sin (2 x) d x

Apply Integration By Parts

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

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

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= \frac{1}{2} [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{2 π}

Simplify

\frac{1}{2} [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{2 π} : \frac{1}{2} [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{2 π}

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

Compute the boundaries:

- π

= \frac{1}{2} (- π)

Simplify

= - \frac{π}{2}

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