What is the integral of cot(10x)csc^4(10x) ?

The integral of cot(10x)csc^4(10x) is -1/40 csc^4(10x)+C

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integral of cot(10x)csc^4(10x)

\int cot (10 x) csc^{4} (10 x) d x

- \frac{1}{40} csc^{4} (10 x) + C

Solution steps

\int cot (10 x) csc^{4} (10 x) d x

Apply u-substitution

: \int cot (u) csc^{4} (u) \frac{1}{10} d u

= \int cot (u) csc^{4} (u) \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int cot (u) csc^{4} (u) d u

Rewrite using trig identities

= \frac{1}{10} \cdot \int \frac{cos ( u )}{sin ^{5} ( u )} d u

Apply u-substitution

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

= \frac{1}{10} \cdot \int \frac{1}{v ^{5}} d v

Apply the Power Rule

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

= \frac{1}{10} (- \frac{1}{4 v ^{4}})

Substitute back

= \frac{1}{10} (- \frac{1}{4 sin ^{4} ( 10 x )})

Simplify

\frac{1}{10} (- \frac{1}{4 sin ^{4} ( 10 x )}) : - \frac{1}{40} csc^{4} (10 x)

= - \frac{1}{40} csc^{4} (10 x)

Add a constant to the solution

= - \frac{1}{40} csc^{4} (10 x) + C

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 5)^{7})

\frac{\partial}{\partial x} (2 x + y)

\frac{\partial}{\partial x} (4 (y + 3)^{3} (x - 4))

x \to - 8 lim (∣ x ∣)

\frac{d y}{d x} = (sin (5 x))

What is the integral of cot(10x)csc^4(10x) ?
The integral of cot(10x)csc^4(10x) is -1/40 csc^4(10x)+C