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

\int e^{3 x + e^{x}} d x

6 (t - 10)

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