What is the integral of cot^3(3x)csc^4(3x) ?

The integral of cot^3(3x)csc^4(3x) is 1/3 (-1/6 csc^6(3x)+1/4 csc^4(3x))+C

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integral of cot^3(3x)csc^4(3x)

\int cot^{3} (3 x) csc^{4} (3 x) d x

\frac{1}{3} (- \frac{1}{6} csc^{6} (3 x) + \frac{1}{4} csc^{4} (3 x)) + C

Solution steps

\int cot^{3} (3 x) csc^{4} (3 x) d x

Apply u-substitution

: \int cot^{3} (u) csc^{4} (u) \frac{1}{3} d u

= \int cot^{3} (u) csc^{4} (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^{4} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{7} ( u )} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{7}} d v

= \frac{1}{3} \cdot \int \frac{1 - v ^{2}}{v ^{7}} d v

Expand

\frac{1 - v ^{2}}{v ^{7}} : \frac{1}{v ^{7}} - \frac{1}{v ^{5}}

= \frac{1}{3} \cdot \int \frac{1}{v ^{7}} - \frac{1}{v ^{5}} 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}{3} (\int \frac{1}{v ^{7}} d v - \int \frac{1}{v ^{5}} d v)

\int \frac{1}{v ^{7}} d v = - \frac{1}{6 v ^{6}}

\int \frac{1}{v ^{5}} d v = - \frac{1}{4 v ^{4}}

= \frac{1}{3} (- \frac{1}{6 v ^{6}} - (- \frac{1}{4 v ^{4}}))

Substitute back

= \frac{1}{3} (- \frac{1}{6 sin ^{6} ( 3 x )} - (- \frac{1}{4 sin ^{4} ( 3 x )}))

Simplify

\frac{1}{3} (- \frac{1}{6 sin ^{6} ( 3 x )} - (- \frac{1}{4 sin ^{4} ( 3 x )})) : \frac{1}{3} (- \frac{1}{6} csc^{6} (3 x) + \frac{1}{4} csc^{4} (3 x))

= \frac{1}{3} (- \frac{1}{6} csc^{6} (3 x) + \frac{1}{4} csc^{4} (3 x))

Add a constant to the solution

= \frac{1}{3} (- \frac{1}{6} csc^{6} (3 x) + \frac{1}{4} csc^{4} (3 x)) + C

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What is the integral of cot^3(3x)csc^4(3x) ?
The integral of cot^3(3x)csc^4(3x) is 1/3 (-1/6 csc^6(3x)+1/4 csc^4(3x))+C