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

The integral of (t^3+2t)sin(t) is -t^3cos(t)+3t^2sin(t)+4tcos(t)-4sin(t)+C

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

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

- t^{3} cos (t) + 3 t^{2} sin (t) + 4 t cos (t) - 4 sin (t) + C

Solution steps

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

Expand

(t^{3} + 2 t) sin (t) : t^{3} sin (t) + 2 t sin (t)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int t^{3} sin (t) d t = - t^{3} cos (t) + 3 (t^{2} sin (t) - 2 (- t cos (t) + sin (t)))

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

= - t^{3} cos (t) + 3 (t^{2} sin (t) - 2 (- t cos (t) + sin (t))) + 2 (- t cos (t) + sin (t))

Simplify

- t^{3} cos (t) + 3 (t^{2} sin (t) - 2 (- t cos (t) + sin (t))) + 2 (- t cos (t) + sin (t)) : - t^{3} cos (t) + 3 t^{2} sin (t) + 4 t cos (t) - 4 sin (t)

= - t^{3} cos (t) + 3 t^{2} sin (t) + 4 t cos (t) - 4 sin (t)

Add a constant to the solution

= - t^{3} cos (t) + 3 t^{2} sin (t) + 4 t cos (t) - 4 sin (t) + C

\frac{d}{d x} (\frac{a - x}{a + x})

f (x) = ln (1 + e^{x}), at x = ln (2)

\int x (x - 3)^{2} d x

y = x^{4} - 4 x^{2} + 4, y = x^{2}

g (t) = 2 t^{\frac{- 3}{4}}

What is the integral of (t^3+2t)sin(t) ?
The integral of (t^3+2t)sin(t) is -t^3cos(t)+3t^2sin(t)+4tcos(t)-4sin(t)+C