What is the integral from pi to 2pi of rsin(r) ?

The integral from pi to 2pi of rsin(r) is -3pi

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integral from pi to 2pi of rsin(r)

\int_{π}^{2 π} r sin (r) d r

- 3 π

Solution steps

\int_{π}^{2 π} r sin (r) d r

Apply Integration By Parts

: [- r cos (r) - \int - cos (r) d r]_{π}^{2 π}

= [- r cos (r) - \int - cos (r) d r]_{π}^{2 π}

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

= [- r cos (r) - (- sin (r))]_{π}^{2 π}

Simplify

= [- r cos (r) + sin (r)]_{π}^{2 π}

Compute the boundaries:

- 3 π

= - 3 π

\int_{\frac{- π}{2}}^{\frac{π}{2}} g (x) d x

\int_{0}^{1} (\frac{x - 1}{x + 1})^{3} d x

\int_{- 2}^{1} (3 x^{2} - 6 x + 2) d x

\int_{- 3}^{1} \frac{1}{( x + 1 ) ^{2}} d x

\int_{0}^{y} 9 + y^{2} d y

What is the integral from pi to 2pi of rsin(r) ?
The integral from pi to 2pi of rsin(r) is -3pi