What is the integral of sin(0.01t)+t^3 ?

The integral of sin(0.01t)+t^3 is -100cos(0.01t)+0.25t^4+C

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integral of sin(0.01t)+t^3

\int sin (0.01 t) + t^{3} d t

- 100 cos (0.01 t) + 0.25 t^{4} + C

Solution steps

\int sin (0.01 t) + t^{3} 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 sin (0.01 t) d t + \int t^{3} d t

\int sin (0.01 t) d t = - 100 cos (0.01 t)

\int t^{3} d t = \frac{t ^{4}}{4}

= - 100 cos (0.01 t) + \frac{t ^{4}}{4}

Simplify

= - 100 cos (0.01 t) + 0.25 t^{4}

Add a constant to the solution

= - 100 cos (0.01 t) + 0.25 t^{4} + C

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What is the integral of sin(0.01t)+t^3 ?
The integral of sin(0.01t)+t^3 is -100cos(0.01t)+0.25t^4+C