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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{v ^{5}} d v

= \frac{1}{3} \cdot \int \frac{1}{v ^{5}} d v

Apply the Power Rule

: - \frac{1}{4 v ^{4}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 45

f (x) = 3 x^{4}, at x = - 2, 48

(e^{- 2 t} - 1)^{2}

(1 - x^{2}) y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

2^{s i n (x)}

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