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

The integral of (sin(3x))/((1-cos(3x))^3) is -1/(6(1-cos(3x))^2)+C

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integral of (sin(3x))/((1-cos(3x))^3)

\int \frac{sin ( 3 x )}{( 1 - cos ( 3 x ) ) ^{3}} d x

- \frac{1}{6 ( 1 - cos ( 3 x ) ) ^{2}} + C

Solution steps

\int \frac{sin ( 3 x )}{( 1 - cos ( 3 x ) ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{3}} d u

= \int \frac{1}{3 u ^{3}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = 1 - cos (3 x)

= \frac{1}{3} (- \frac{1}{2 ( 1 - cos ( 3 x ) ) ^{2}})

Simplify

\frac{1}{3} (- \frac{1}{2 ( 1 - cos ( 3 x ) ) ^{2}}) : - \frac{1}{6 ( 1 - cos ( 3 x ) ) ^{2}}

= - \frac{1}{6 ( 1 - cos ( 3 x ) ) ^{2}}

Add a constant to the solution

= - \frac{1}{6 ( 1 - cos ( 3 x ) ) ^{2}} + C

\int \frac{1}{2 ^{2} + u ^{2}} d u

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\int_{0}^{4} 8 x d x

\int \frac{x + 1}{x ^{2} + 2 x} d x

\frac{d y}{d x} = cos (2 x)

What is the integral of (sin(3x))/((1-cos(3x))^3) ?
The integral of (sin(3x))/((1-cos(3x))^3) is -1/(6(1-cos(3x))^2)+C