What is the integral of sin(8x-2) ?

The integral of sin(8x-2) is -1/8 cos(8x-2)+C

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integral of sin(8x-2)

\int sin (8 x - 2) d x

- \frac{1}{8} cos (8 x - 2) + C

Solution steps

\int sin (8 x - 2) d x

Apply u-substitution

: \int sin (u) \frac{1}{8} d u

= \int sin (u) \frac{1}{8} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{8} (- cos (u))

Substitute back

u = 8 x - 2

= \frac{1}{8} (- cos (8 x - 2))

Simplify

= - \frac{1}{8} cos (8 x - 2)

Add a constant to the solution

= - \frac{1}{8} cos (8 x - 2) + C

f^{^{'}} (x) = 6 x^{2} - 8 x, f (3) = 27

x \to - 9 lim (\frac{- 3 x}{x + 9})

\frac{d}{d t} (a sin^{3} (t))

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

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

What is the integral of sin(8x-2) ?
The integral of sin(8x-2) is -1/8 cos(8x-2)+C