What is the integral of cos(9x) ?

The integral of cos(9x) is 1/9 sin(9x)+C

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integral of cos(9x)

\int cos (9 x) d x

\frac{1}{9} sin (9 x) + C

Solution steps

\int cos (9 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{9} d u

= \int cos (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{9} sin (u)

Substitute back

u = 9 x

= \frac{1}{9} sin (9 x)

Add a constant to the solution

= \frac{1}{9} sin (9 x) + C

\int cot (\frac{1}{8} x) d x

\frac{d}{d x} (\frac{4 x ^{\frac{3}{2}} - x ^{3} + 8}{x})

\frac{d}{d x} (\frac{1}{2 x - 1})

\frac{d e}{d t}

\frac{2 - 5 s}{s ^{2} + 9}