What is the integral of (\sqrt[3]{t^2}-sin(t)) ?

The integral of (\sqrt[3]{t^2}-sin(t)) is 3/5 t^{5/3}+cos(t)+C

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integral of (\sqrt[3]{t^2}-sin(t))

\frac{3}{5} t^{\frac{5}{3}} + cos (t) + C

Solution steps

Apply the Sum Rule:

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

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

= \frac{3}{5} t^{\frac{5}{3}} - (- cos (t))

Simplify

= \frac{3}{5} t^{\frac{5}{3}} + cos (t)

Add a constant to the solution

= \frac{3}{5} t^{\frac{5}{3}} + cos (t) + C

\frac{\partial}{\partial x} (100 x^{0.66} y^{0.34})

\frac{d}{d x} (a^{3 x^{2}})

- 4 x^{3} + 3 x^{2} - 6 x + 10

y = x^{3} - 14 x^{2} + 32 x + 4

s \to 0 lim (\frac{- 2 s}{s - 3})

What is the integral of (\sqrt[3]{t^2}-sin(t)) ?
The integral of (\sqrt[3]{t^2}-sin(t)) is 3/5 t^{5/3}+cos(t)+C