What is the integral of 8sin(2x+90) ?

The integral of 8sin(2x+90) is -4cos(2x+90)+C

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integral of 8sin(2x+90)

\int 8 sin (2 x + 90) d x

- 4 cos (2 x + 90) + C

Solution steps

\int 8 sin (2 x + 90) d x

Take the constant out:

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

= 8 \cdot \int sin (2 x + 90) d x

Apply u-substitution

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

= 8 \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

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

Use the common integral:

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

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

Substitute back

u = 2 x + 90

= 8 \cdot \frac{1}{2} (- cos (2 x + 90))

Simplify

8 \cdot \frac{1}{2} (- cos (2 x + 90)) : - 4 cos (2 x + 90)

= - 4 cos (2 x + 90)

Add a constant to the solution

= - 4 cos (2 x + 90) + C

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What is the integral of 8sin(2x+90) ?
The integral of 8sin(2x+90) is -4cos(2x+90)+C