What is the integral of 8sin(4y) ?

The integral of 8sin(4y) is -2cos(4y)+C

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integral of 8sin(4y)

\int 8 sin (4 y) d y

- 2 cos (4 y) + C

Solution steps

\int 8 sin (4 y) d y

Take the constant out:

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

= 8 \cdot \int sin (4 y) d y

Apply u-substitution

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

= 8 \cdot \int sin (u) \frac{1}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int sin (u) d u

Use the common integral:

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

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

Substitute back

u = 4 y

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

Simplify

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

= - 2 cos (4 y)

Add a constant to the solution

= - 2 cos (4 y) + C

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