What is the integral of-1/8 sin(2x) ?

The integral of-1/8 sin(2x) is 1/16 cos(2x)+C

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integral of-1/8 sin(2x)

\int - \frac{1}{8} sin (2 x) d x

\frac{1}{16} cos (2 x) + C

Solution steps

\int - \frac{1}{8} sin (2 x) d x

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

- \frac{1}{8} \cdot \frac{1}{2} (- cos (2 x)) : \frac{1}{16} cos (2 x)

= \frac{1}{16} cos (2 x)

Add a constant to the solution

= \frac{1}{16} cos (2 x) + C

x \to 1 lim (lo g_{4} (x))

x \to 8 + lim (\frac{1}{x - 8} - \frac{1}{x - 8})

x \to 4 lim (\frac{5 + x}{x - 4})

n = 1 \sum \infty \frac{2}{n ^{3}}

\int t 7 t^{2} + 12 d t

What is the integral of-1/8 sin(2x) ?
The integral of-1/8 sin(2x) is 1/16 cos(2x)+C