What is the integral of cot^3(3x)csc^2(3x) ?

The integral of cot^3(3x)csc^2(3x) is -1/12 cot^4(3x)+C

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integral of cot^3(3x)csc^2(3x)

\int cot^{3} (3 x) csc^{2} (3 x) d x

- \frac{1}{12} cot^{4} (3 x) + C

Solution steps

\int cot^{3} (3 x) csc^{2} (3 x) d x

Apply u-substitution

: \int cot^{3} (u) csc^{2} (u) \frac{1}{3} d u

= \int cot^{3} (u) csc^{2} (u) \frac{1}{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 cot^{3} (u) csc^{2} (u) d u

Apply u-substitution

: \int - v^{3} d v

= \frac{1}{3} \cdot \int - v^{3} d v

Take the constant out:

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

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

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{3} (- \frac{v ^{4}}{4})

Substitute back

= \frac{1}{3} (- \frac{cot ^{4} ( 3 x )}{4})

Simplify

\frac{1}{3} (- \frac{cot ^{4} ( 3 x )}{4}) : - \frac{1}{12} cot^{4} (3 x)

= - \frac{1}{12} cot^{4} (3 x)

Add a constant to the solution

= - \frac{1}{12} cot^{4} (3 x) + C

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What is the integral of cot^3(3x)csc^2(3x) ?
The integral of cot^3(3x)csc^2(3x) is -1/12 cot^4(3x)+C