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

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

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integral of-3t^2cos(t)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{- 3}{2 ^{n}}

x, x^{2}, - x + 1

\frac{d}{d x} (ln (\frac{x - 1}{x + 1}))

\frac{\partial}{\partial x} (\frac{m}{x ^{a} y ^{b} z ^{c}})

\frac{\partial}{\partial x} (\frac{3}{8} x y)

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