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

The integral of (t^3)/m-(cos(t))/m is -1/m sin(t)+(t^4)/(4m)+C

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integral of (t^3)/m-(cos(t))/m

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

- \frac{1}{m} sin (t) + \frac{t ^{4}}{4 m} + C

Solution steps

\int \frac{t ^{3}}{m} - \frac{cos ( t )}{m} 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 \frac{cos ( t )}{m} d t + \int \frac{t ^{3}}{m} d t

\int \frac{cos ( t )}{m} d t = \frac{1}{m} sin (t)

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

= - \frac{1}{m} sin (t) + \frac{t ^{4}}{4 m}

Add a constant to the solution

= - \frac{1}{m} sin (t) + \frac{t ^{4}}{4 m} + C

\frac{\partial}{\partial x} (y^{2} ln (y) - 2 x y^{2})

\frac{d}{d x} (\frac{x ^{2} - 9 x + 20}{( x + 2 ) ( x - 2 )})

\frac{d}{d x} ((lo g_{10} (x))^{s i n h (x)})

f (x) = x^{2} - 4, at a = 1

\int_{5}^{9} (9 - y + 2)^{2} d y

What is the integral of (t^3)/m-(cos(t))/m ?
The integral of (t^3)/m-(cos(t))/m is -1/m sin(t)+(t^4)/(4m)+C