What is the integral of t^2sin(4t) ?

The integral of t^2sin(4t) is -1/4 t^2cos(4t)+1/32 (4tsin(4t)+cos(4t))+C

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integral of t^2sin(4t)

\int t^{2} sin (4 t) d t

- \frac{1}{4} t^{2} cos (4 t) + \frac{1}{32} (4 t sin (4 t) + cos (4 t)) + C

Solution steps

\int t^{2} sin (4 t) d t

Apply Integration By Parts

= - \frac{1}{4} t^{2} cos (4 t) - \int - \frac{1}{2} t cos (4 t) d t

\int - \frac{1}{2} t cos (4 t) d t = - \frac{1}{32} (4 t sin (4 t) + cos (4 t))

= - \frac{1}{4} t^{2} cos (4 t) - (- \frac{1}{32} (4 t sin (4 t) + cos (4 t)))

Simplify

= - \frac{1}{4} t^{2} cos (4 t) + \frac{1}{32} (4 t sin (4 t) + cos (4 t))

Add a constant to the solution

= - \frac{1}{4} t^{2} cos (4 t) + \frac{1}{32} (4 t sin (4 t) + cos (4 t)) + C

What is the integral of t^2sin(4t) ?
The integral of t^2sin(4t) is -1/4 t^2cos(4t)+1/32 (4tsin(4t)+cos(4t))+C