What is the integral of cos(b+ax) ?

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

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integral of cos(b+ax)

\int cos (b + a x) d x

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

Solution steps

\int cos (b + a x) d x

Apply u-substitution

: \int \frac{cos ( u )}{a} d u

= \int \frac{cos ( u )}{a} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{a} sin (u)

Substitute back

u = b + a x

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

Add a constant to the solution

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

2 x^{2} y y^{^{'}} + 2 x y^{2} + 1 = 0

\int (4^{x}) d x

\frac{d}{d x} (3 ln (x + 1))

\frac{d}{d x} (35 x ln (13 x))

\frac{\partial}{\partial x} (- 3 x y^{2})

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