What is the integral from 2 to y of t^2sin(t) ?

The integral from 2 to y of t^2sin(t) is -y^2cos(y)+2ysin(y)+2cos(y)+2cos(2)-4sin(2)

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integral from 2 to y of t^2sin(t)

\int_{2}^{y} t^{2} sin (t) d t

- y^{2} cos (y) + 2 y sin (y) + 2 cos (y) + 2 cos (2) - 4 sin (2)

Solution steps

\int_{2}^{y} t^{2} sin (t) d t

Apply Integration By Parts

: [- t^{2} cos (t) - \int - 2 t cos (t) d t]_{2}^{y}

= [- t^{2} cos (t) - \int - 2 t cos (t) d t]_{2}^{y}

\int - 2 t cos (t) d t = - 2 (t sin (t) + cos (t))

= [- t^{2} cos (t) - (- 2 (t sin (t) + cos (t)))]_{2}^{y}

Simplify

= [- t^{2} cos (t) + 2 (t sin (t) + cos (t))]_{2}^{y}

Compute the boundaries:

- y^{2} cos (y) + 2 y sin (y) + 2 cos (y) + 2 cos (2) - 4 sin (2)

= - y^{2} cos (y) + 2 y sin (y) + 2 cos (y) + 2 cos (2) - 4 sin (2)

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What is the integral from 2 to y of t^2sin(t) ?
The integral from 2 to y of t^2sin(t) is -y^2cos(y)+2ysin(y)+2cos(y)+2cos(2)-4sin(2)