What is the integral of 5sin^3(9x) ?

The integral of 5sin^3(9x) is 5/9 (-cos(9x)+1/3 cos^3(9x))+C

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integral of 5sin^3(9x)

\int 5 sin^{3} (9 x) d x

\frac{5}{9} (- cos (9 x) + \frac{1}{3} cos^{3} (9 x)) + C

Solution steps

\int 5 sin^{3} (9 x) d x

Take the constant out:

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

= 5 \cdot \int sin^{3} (9 x) d x

Apply u-substitution

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

= 5 \cdot \int sin^{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

= 5 \cdot \frac{1}{9} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{9} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \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}

= 5 \cdot \frac{1}{9} (- v + \frac{v ^{3}}{3})

Substitute back

= 5 \cdot \frac{1}{9} (- cos (9 x) + \frac{cos ^{3} ( 9 x )}{3})

Simplify

5 \cdot \frac{1}{9} (- cos (9 x) + \frac{cos ^{3} ( 9 x )}{3}) : \frac{5}{9} (- cos (9 x) + \frac{1}{3} cos^{3} (9 x))

= \frac{5}{9} (- cos (9 x) + \frac{1}{3} cos^{3} (9 x))

Add a constant to the solution

= \frac{5}{9} (- cos (9 x) + \frac{1}{3} cos^{3} (9 x)) + C

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What is the integral of 5sin^3(9x) ?
The integral of 5sin^3(9x) is 5/9 (-cos(9x)+1/3 cos^3(9x))+C