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

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

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integral of cos(ax+b)

\int cos (a x + b) d x

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

Solution steps

\int cos (a x + b) 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 = a x + b

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

Add a constant to the solution

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

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x \to - 3 lim (3 x)

\int 5 x + 3 d x

\frac{\partial}{\partial x} (d y)

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