What is the integral of 6cos(3x) ?

The integral of 6cos(3x) is 2sin(3x)+C

integral of 6cos(3x)

\int 6 cos (3 x) d x

2 sin (3 x) + C

Solution steps

\int 6 cos (3 x) d x

Take the constant out:

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

= 6 \cdot \int cos (3 x) d x

Apply u-substitution

= 6 \cdot \int cos (u) \frac{1}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} sin (u)

Substitute back

u = 3 x

= 6 \cdot \frac{1}{3} sin (3 x)

Simplify

6 \cdot \frac{1}{3} sin (3 x) : 2 sin (3 x)

= 2 sin (3 x)

Add a constant to the solution

= 2 sin (3 x) + C

\frac{d}{d x} (a cos (2 x) + b sin (2 x))

\frac{d}{d x} (a sin (x))

\int tan (2 x se) c^{2} 2 x d x

t an g e n t y = 3 x - 2 x^{2}, (- 3, - 27)

x \to 0 + lim (x + 1)