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

The integral from 0 to pi of tcos(t) is -2

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integral from 0 to pi of tcos(t)

\int_{0}^{π} t cos (t) d t

- 2

Solution steps

\int_{0}^{π} t cos (t) d t

Apply Integration By Parts

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

= [t sin (t) - \int sin (t) d t]_{0}^{π}

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

= [t sin (t) - (- cos (t))]_{0}^{π}

Simplify

= [t sin (t) + cos (t)]_{0}^{π}

Compute the boundaries:

- 2

= - 2

\int_{0}^{2} \frac{1}{x ^{2} + 6 x + 13} d x

\int_{- 1}^{2} (- 2 x^{2} + 2 x + 4) d x

\int_{0}^{\frac{π}{5}} t^{2} sin (5 t) d t

\int_{1}^{2} (2 x + 1)^{2} d x

\int_{0}^{2} (- 2 x^{2} + 4 x) d x

What is the integral from 0 to pi of tcos(t) ?
The integral from 0 to pi of tcos(t) is -2