What is the integral of sin(a+bx) ?

The integral of sin(a+bx) is -1/b cos(a+bx)+C

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integral of sin(a+bx)

\int sin (a + b x) d x

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

Solution steps

\int sin (a + b x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = a + b x

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (2 x ln (3 x) + x)

\frac{\partial}{\partial x} (x + 2 y - 22)

x \to - \infty lim (ln ∣ x ∣)

\frac{\partial}{\partial x} (7 e^{y} - 4 y e^{x})

sin (9 x)

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