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

\int (14 x + 6 x) d x

y^{^{'}} + y = y x e^{x + 2}

\int \frac{x}{3 x + 4} d x

x \to 3 lim (\frac{1}{( x - 3 ) ^{8}})

x sin (\frac{3}{x})

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