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

The integral of t^3sin(2t^2) is 1/8 (-2t^2cos(2t^2)+sin(2t^2))+C

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

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

\frac{1}{8} (- 2 t^{2} cos (2 t^{2}) + sin (2 t^{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{u sin ( u )}{8} d u

= \int \frac{u sin ( u )}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int u sin (u) d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= \frac{1}{8} (- u cos (u) - \int - cos (u) d u)

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

= \frac{1}{8} (- u cos (u) - (- sin (u)))

Substitute back

u = 2 t^{2}

= \frac{1}{8} (- 2 t^{2} cos (2 t^{2}) - (- sin (2 t^{2})))

Simplify

= \frac{1}{8} (- 2 t^{2} cos (2 t^{2}) + sin (2 t^{2}))

Add a constant to the solution

= \frac{1}{8} (- 2 t^{2} cos (2 t^{2}) + sin (2 t^{2})) + C

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What is the integral of t^3sin(2t^2) ?
The integral of t^3sin(2t^2) is 1/8 (-2t^2cos(2t^2)+sin(2t^2))+C