What is the integral of cos^3(9x) ?

The integral of cos^3(9x) is 1/9 (sin(9x)-(sin^3(9x))/3)+C

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integral of cos^3(9x)

\int cos^{3} (9 x) d x

\frac{1}{9} (sin (9 x) - \frac{sin ^{3} ( 9 x )}{3}) + C

Solution steps

\int cos^{3} (9 x) d x

Apply u-substitution

: \int cos^{3} (u) \frac{1}{9} d u

= \int cos^{3} (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^{3} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= \frac{1}{9} \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{9} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{9} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{9} (sin (9 x) - \frac{sin ^{3} ( 9 x )}{3})

Add a constant to the solution

= \frac{1}{9} (sin (9 x) - \frac{sin ^{3} ( 9 x )}{3}) + C

\int_{0}^{1} \frac{1}{( x ^{2} + 1 )} d x

\int \frac{cos ( \frac{π}{x ^{41}} )}{x ^{42}} d x

\int ln (y^{2}) d y

x \to 1 - lim (\frac{2 - x}{x - 1})

What is the integral of cos^3(9x) ?
The integral of cos^3(9x) is 1/9 (sin(9x)-(sin^3(9x))/3)+C