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

The integral of-9cos(t) is -9sin(t)+C

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integral of-9cos(t)

\int - 9 cos (t) d t

- 9 sin (t) + C

Solution steps

\int - 9 cos (t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= - 9 sin (t)

Add a constant to the solution

= - 9 sin (t) + C

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