What is the integral of 6cos(2x) ?

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

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integral of 6cos(2x)

\int 6 cos (2 x) d x

3 sin (2 x) + C

Solution steps

\int 6 cos (2 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 (2 x) d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} sin (u)

Substitute back

u = 2 x

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

Simplify

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

= 3 sin (2 x)

Add a constant to the solution

= 3 sin (2 x) + C

\frac{\partial}{\partial x} (x^{2} - 1)

\frac{d}{d x} ((x - 3 x^{4}) tan (x))

3 ln (5 x)

\frac{d y}{d x} = x \cdot 1 - y^{2}

\frac{d}{d x} (\frac{1 - x ^{3}}{1 + x ^{3}})