What is the integral of-4sin(-2t) ?

The integral of-4sin(-2t) is -2cos(2t)+C

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integral of-4sin(-2t)

\int - 4 sin (- 2 t) d t

- 2 cos (2 t) + C

Solution steps

\int - 4 sin (- 2 t) d t

Take the constant out:

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

= - 4 \cdot \int sin (- 2 t) d t

Apply u-substitution

: \int - \frac{1}{2} sin (u) d u

= - 4 \cdot \int - \frac{1}{2} sin (u) d u

Take the constant out:

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

= - 4 (- \frac{1}{2} \cdot \int sin (u) d u)

Use the common integral:

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

= - 4 (- \frac{1}{2} (- cos (u)))

Substitute back

u = - 2 t

= - 4 (- \frac{1}{2} (- cos (- 2 t)))

Simplify

- 4 (- \frac{1}{2} (- cos (- 2 t))) : - 2 cos (2 t)

= - 2 cos (2 t)

Add a constant to the solution

= - 2 cos (2 t) + C

\int 3 t^{2} (t^{3} + 4)^{5} d t

\frac{1}{1 + \frac{1}{x}}, 0

\int_{- 1}^{1} (x - 1 - x^{2}) d x

t y^{^{'}} = 3 y + t^{4} sin (4 t)

P^{^{'}} = 200 e^{0.02 t}