What is the integral of sin(4x-9) ?

The integral of sin(4x-9) is -1/4 cos(4x-9)+C

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integral of sin(4x-9)

\int sin (4 x - 9) d x

- \frac{1}{4} cos (4 x - 9) + C

Solution steps

\int sin (4 x - 9) d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 4 x - 9

= \frac{1}{4} (- cos (4 x - 9))

Simplify

= - \frac{1}{4} cos (4 x - 9)

Add a constant to the solution

= - \frac{1}{4} cos (4 x - 9) + C

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What is the integral of sin(4x-9) ?
The integral of sin(4x-9) is -1/4 cos(4x-9)+C