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

\int \frac{e ^{2 t}}{e ^{2 t} - 1} d t

\frac{d}{d x} (ln (\frac{1 + tan ( x )}{1 + sin ( x )}))

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{19}{2}}} d t

f (x) = x^{3} sin (x)

x \to \infty lim (x b)

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