What is the integral of (sin(2x))/4 ?

The integral of (sin(2x))/4 is -1/8 cos(2x)+C

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integral of (sin(2x))/4

\int \frac{sin ( 2 x )}{4} d x

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

Solution steps

\int \frac{sin ( 2 x )}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int sin (2 x) d x

Apply u-substitution

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

= \frac{1}{4} \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}{4} \cdot \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{4} \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 x

= \frac{1}{4} \cdot \frac{1}{2} (- cos (2 x))

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x ^{\frac{4}{3}}} d x

\frac{\partial}{\partial y} (x^{2} - y^{3} - x^{2} y + y)

\int_{1}^{R} \frac{ln ( 4 x )}{x ^{2}} d x

\int \frac{e}{π} d x

\frac{\partial}{\partial x} (x^{2} e^{4 x y})

What is the integral of (sin(2x))/4 ?
The integral of (sin(2x))/4 is -1/8 cos(2x)+C