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

y^{^{''}} + y^{^{'}} + y = cos (t)

\int \frac{7 x ^{6} + 2}{x ^{7} + 2 x} d x

2 x^{2}

\int_{- 2}^{- 1} (u - \frac{1}{u ^{2}}) d u

1 + x^{2}

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