What is the integral from 0 to x of sin(4x-y) ?

The integral from 0 to x of sin(4x-y) is cos(3x)-cos(4x)

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integral from 0 to x of sin(4x-y)

\int_{0}^{x} sin (4 x - y) d y

cos (3 x) - cos (4 x)

Solution steps

\int_{0}^{x} sin (4 x - y) d y

Apply u-substitution

: \int_{4 x}^{3 x} - sin (u) d u

= \int_{4 x}^{3 x} - sin (u) d u

Take the constant out:

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

= - \int_{4 x}^{3 x} sin (u) d u

Use the common integral:

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

= - [- cos (u)]_{4 x}^{3 x}

Compute the boundaries:

- cos (3 x) + cos (4 x)

= - (- cos (3 x) + cos (4 x))

Simplify

= cos (3 x) - cos (4 x)

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What is the integral from 0 to x of sin(4x-y) ?
The integral from 0 to x of sin(4x-y) is cos(3x)-cos(4x)