What is the integral from 0 to t of t^2sin(t) ?

The integral from 0 to t of t^2sin(t) is -t^2cos(t)+2(tsin(t)+cos(t))-2

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integral from 0 to t of t^2sin(t)

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

- t^{2} cos (t) + 2 (t sin (t) + cos (t)) - 2

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

- t^{2} cos (t) + 2 (t sin (t) + cos (t)) - 2

= - t^{2} cos (t) + 2 (t sin (t) + cos (t)) - 2

\int_{0}^{1} x^{2} \cdot 3 x^{2} d x

\int_{- 7}^{0} (3 + 49 - x^{2}) d x

\int_{0}^{1} \frac{3 x}{( 2 x + 1 ) ^{3}} d x

\int_{- i}^{3 i} sin (t) d t

\int_{1}^{x^{2}} 16 x y^{3} d y

What is the integral from 0 to t of t^2sin(t) ?
The integral from 0 to t of t^2sin(t) is -t^2cos(t)+2(tsin(t)+cos(t))-2