What is the integral of-16cos(4t+pi) ?

The integral of-16cos(4t+pi) is 4sin(4t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-16cos(4t+pi)

\int - 16 cos (4 t + π) d t

4 sin (4 t) + C

Solution steps

\int - 16 cos (4 t + π) d t

Take the constant out:

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

= - 16 \cdot \int cos (4 t + π) d t

Rewrite using trig identities

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

Take the constant out:

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

= - 16 (- \int cos (4 t) d t)

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

= - 16 (- \int cos (u) \frac{1}{4} d u)

Take the constant out:

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

= - 16 (- \frac{1}{4} \cdot \int cos (u) d u)

Use the common integral:

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

= - 16 (- \frac{1}{4} sin (u))

Substitute back

u = 4 t

= - 16 (- \frac{1}{4} sin (4 t))

Simplify

- 16 (- \frac{1}{4} sin (4 t)) : 4 sin (4 t)

= 4 sin (4 t)

Add a constant to the solution

= 4 sin (4 t) + C

\frac{d}{d x} ((x - 2)^{x + 1})

(y^{- 2})^{^{'}}

\frac{\partial}{\partial x} (x^{6} y - 2 x^{3} y^{2})

y = x + 4, y = x^{2} + x - 12

2 (x + 1), 3 (x + 1), x = 4

What is the integral of-16cos(4t+pi) ?
The integral of-16cos(4t+pi) is 4sin(4t)+C