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

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

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integral of 1/2 sin(8-4x)

\int \frac{1}{2} sin (8 - 4 x) d x

\frac{1}{8} cos (8 - 4 x) + C

Solution steps

\int \frac{1}{2} sin (8 - 4 x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (8 - 4 x) d x

Apply u-substitution

: \int - \frac{1}{4} sin (u) d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 8 - 4 x

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

Simplify

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

= \frac{1}{8} cos (8 - 4 x)

Add a constant to the solution

= \frac{1}{8} cos (8 - 4 x) + C

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