What is the integral of sin(wt-bx) ?

The integral of sin(wt-bx) is -1/w cos(-bx+wt)+C

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integral of sin(wt-bx)

\int sin (wt - b x) d t

- \frac{1}{w} cos (- b x + wt) + C

Solution steps

\int sin (wt - b x) d t

Apply u-substitution

: \int \frac{sin ( u )}{w} d u

= \int \frac{sin ( u )}{w} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - b x + wt

= \frac{1}{w} (- cos (- b x + wt))

Simplify

= - \frac{1}{w} cos (- b x + wt)

Add a constant to the solution

= - \frac{1}{w} cos (- b x + wt) + C

\frac{\partial}{\partial y} (2 x^{2} + 6 x y + y^{2})

c cos (t) + t^{2} sin (t)

\int (2 + 2 cos (θ))^{2} d θ

\int \frac{1}{4 x ^{2} + 4 x + 10} d x

\frac{d y}{d x} = (3 x^{2} y)

What is the integral of sin(wt-bx) ?
The integral of sin(wt-bx) is -1/w cos(-bx+wt)+C