What is the integral from 0 to pi of tcos(t)sin(t) ?

The integral from 0 to pi of tcos(t)sin(t) is -pi/4

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integral from 0 to pi of tcos(t)sin(t)

\int_{0}^{π} t cos (t) sin (t) d t

- \frac{π}{4}

Decimal

- 0.78539 \dots

Solution steps

\int_{0}^{π} t cos (t) sin (t) d t

Rewrite using trig identities

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

- \frac{π}{2}

= \frac{1}{2} (- \frac{π}{2})

Simplify

= - \frac{π}{4}

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What is the integral from 0 to pi of tcos(t)sin(t) ?
The integral from 0 to pi of tcos(t)sin(t) is -pi/4