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

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

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integral of (t^2-2cos(t))

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

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

Solution steps

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

\int t^{2} d t = \frac{t ^{3}}{3}

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

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

Add a constant to the solution

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

f (x) = 2 x^{2} - 3 x + 1

x \to 0 lim (\frac{100}{x ^{2}})

\frac{d}{d t} (e^{2 t})

\int (x - 6) sin (π x) d x

cos^{3} (t)

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