What is the integral of sqrt(t)-9cos(t) ?

The integral of sqrt(t)-9cos(t) is 2/3 t^{3/2}-9sin(t)+C

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integral of sqrt(t)-9cos(t)

\int t - 9 cos (t) d t

\frac{2}{3} t^{\frac{3}{2}} - 9 sin (t) + C

Solution steps

\int t - 9 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 d t - \int 9 cos (t) d t

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

\int 9 cos (t) d t = 9 sin (t)

= \frac{2}{3} t^{\frac{3}{2}} - 9 sin (t)

Add a constant to the solution

= \frac{2}{3} t^{\frac{3}{2}} - 9 sin (t) + C

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What is the integral of sqrt(t)-9cos(t) ?
The integral of sqrt(t)-9cos(t) is 2/3 t^{3/2}-9sin(t)+C