What is the integral of 8cos(2x) ?

The integral of 8cos(2x) is 4sin(2x)+C

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integral of 8cos(2x)

\int 8 cos (2 x) d x

4 sin (2 x) + C

Solution steps

\int 8 cos (2 x) d x

Take the constant out:

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

= 8 \cdot \int cos (2 x) d x

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} sin (u)

Substitute back

u = 2 x

= 8 \cdot \frac{1}{2} sin (2 x)

Simplify

8 \cdot \frac{1}{2} sin (2 x) : 4 sin (2 x)

= 4 sin (2 x)

Add a constant to the solution

= 4 sin (2 x) + C

\int \frac{1}{( x - 6 ) ( x - 3 )} d x

\frac{\partial}{\partial x} (\frac{93}{2 + x ^{2} + y ^{2}})

x \to 1 lim (\frac{x}{∣ x ∣})

y = (ln (x))^{l n (x)}

\frac{\partial}{\partial x} (\frac{z}{x + y})