What is the integral of (-16cos(2t)) ?

The integral of (-16cos(2t)) is -8sin(2t)+C

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integral of (-16cos(2t))

\int (- 16 cos (2 t)) d t

- 8 sin (2 t) + C

Solution steps

\int - 16 cos (2 t) d t

Take the constant out:

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

= - 16 \cdot \int cos (2 t) d t

Apply u-substitution

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

= - 16 \cdot \int cos (u) \frac{1}{2} d u

Take the constant out:

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

= - 16 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= - 16 \cdot \frac{1}{2} sin (u)

Substitute back

u = 2 t

= - 16 \cdot \frac{1}{2} sin (2 t)

Simplify

- 16 \cdot \frac{1}{2} sin (2 t) : - 8 sin (2 t)

= - 8 sin (2 t)

Add a constant to the solution

= - 8 sin (2 t) + C

\int (\frac{sin ( x )}{cos ( x )}) d x

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

y^{^{'}} = \frac{y ^{2} + 2}{x y - 4 y}

\frac{\partial}{\partial x} (x y e^{4 y})

y = x^{2} + 6 x - 5, at x = 2

What is the integral of (-16cos(2t)) ?
The integral of (-16cos(2t)) is -8sin(2t)+C