What is the integral of 8cos(3x) ?

The integral of 8cos(3x) is 8/3 sin(3x)+C

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integral of 8cos(3x)

\int 8 cos (3 x) d x

\frac{8}{3} sin (3 x) + C

Solution steps

\int 8 cos (3 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 (3 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{3} d u

= 8 \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

= 8 \cdot \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{3} sin (u)

Substitute back

u = 3 x

= 8 \cdot \frac{1}{3} sin (3 x)

Simplify

8 \cdot \frac{1}{3} sin (3 x) : \frac{8}{3} sin (3 x)

= \frac{8}{3} sin (3 x)

Add a constant to the solution

= \frac{8}{3} sin (3 x) + C

4 + 7 x + x^{2} + δ

x \to 0 lim (\frac{7 x ^{2}}{x ^{2}})

x \to 5 lim (75)

\frac{d}{d t} (a (t - sin (t)))

\frac{7}{x}