integral of -16cos(4t+pi) — step-by-step solution

1) Before integrating, let's calculate:

-16 \cdot \cos (4 t + π)

ⓘ A shift by π changes the sign of cos and sin:

\cos (4 t + π)

- \cos (4 t)

-16 \cdot (- \cos (4 t))

ⓘ Two minus signs cancel out:

-16 \cdot (- \cos (4 t))

16 \cdot \cos (4 t)

16 \cdot \cos (4 t)

2) Let's compute the integral:

\int 16 \cdot \cos (4 t) d t

For the math expression:

16 \cdot \cos (4 t)

, since (sin(a))' = a'*cos(a), we try differentiating

\sin (4 t)

3) Let's perform the differentiation:

\frac{d}{d t} (\sin (4 t))

We differentiate using:

\frac{d}{d t} (\sin (f)) = \frac{d}{d t} (f) \cdot \cos (f)

(where f =

4 t

)

ⓘ The derivative of sin(x) is cos(x):

\frac{d}{d t} (\sin (4 t))

\frac{d}{d t} (4 t) \cdot \cos (4 t)

\frac{d}{d t} (4 t) \cdot \cos (4 t)

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d t} (4 t)

4

4 \cdot \cos (4 t)

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⓘ The derivative of

4 \cdot \sin (4 t)

16 \cdot \cos (4 t)

, so

4 \cdot \sin (4 t)

is an antiderivative of

16 \cdot \cos (4 t)

\int 16 \cdot \cos (4 t) d t

4 \cdot \sin (4 t)

4 \cdot \sin (4 t)

▶Indefinite integrals are defined up to an additive constant C, so this yields:

4 \cdot \sin (4 t) + C