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 sin (u) \frac{1}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 t

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

Simplify

- 4 \cdot \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

x \to 0 - lim (\frac{x ^{2}}{7} - \frac{6}{x})

\int_{- 3}^{3} (3 - ∣ x ∣) d x

\frac{d y}{d x} - \frac{2 y}{x} = x^{2}

(2.3) (5.2)

f^{^{'}} (x) = x ∣ x ∣