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

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

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integral of 1/8 cos(2x)

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

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

Solution steps

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

Apply u-substitution

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

= \frac{1}{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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

(1 + x^{2}) \frac{d y}{d x} - 2 x y = 0

\frac{d y}{d x} = \frac{3 x + x y ^{2}}{2 y + x ^{2} y}

\frac{d}{d x} (x (ln (5 x))^{2})

\frac{d y}{d t} = (k (y - 60))

y^{^{'}} - 3 y = e^{3 x} 4 - x^{2}

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