What is the integral of (2t-t^2)sin(t^2-(t^3)/3) ?

The integral of (2t-t^2)sin(t^2-(t^3)/3) is -cos(t^2-(t^3)/3)+C

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integral of (2t-t^2)sin(t^2-(t^3)/3)

\int (2 t - t^{2}) sin (t^{2} - \frac{t ^{3}}{3}) d t

- cos (t^{2} - \frac{t ^{3}}{3}) + C

Solution steps

\int (2 t - t^{2}) sin (t^{2} - \frac{t ^{3}}{3}) d t

Apply u-substitution

: \int sin (u) d u

= \int sin (u) d u

Use the common integral:

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

= - cos (u)

Substitute back

u = t^{2} - \frac{t ^{3}}{3}

= - cos (t^{2} - \frac{t ^{3}}{3})

Add a constant to the solution

= - cos (t^{2} - \frac{t ^{3}}{3}) + C

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What is the integral of (2t-t^2)sin(t^2-(t^3)/3) ?
The integral of (2t-t^2)sin(t^2-(t^3)/3) is -cos(t^2-(t^3)/3)+C