What is the integral of-9sin(3t) ?

The integral of-9sin(3t) is 3cos(3t)+C

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integral of-9sin(3t)

\int - 9 sin (3 t) d t

3 cos (3 t) + C

Solution steps

\int - 9 sin (3 t) d t

Take the constant out:

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

= - 9 \cdot \int sin (3 t) d t

Apply u-substitution

: \int sin (u) \frac{1}{3} d u

= - 9 \cdot \int sin (u) \frac{1}{3} d u

Take the constant out:

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

= - 9 \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= - 9 \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 t

= - 9 \cdot \frac{1}{3} (- cos (3 t))

Simplify

- 9 \cdot \frac{1}{3} (- cos (3 t)) : 3 cos (3 t)

= 3 cos (3 t)

Add a constant to the solution

= 3 cos (3 t) + C

x \to 1 lim (x^{17} - x + 3)

\frac{d}{d x} (3 x^{2} + 2 y^{4})

\frac{d}{d y} (e^{- x y})

\int (3 - x^{2}) 3 d x