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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{cos ( x )}{sin ^{5} ( x )} d x

Apply u-substitution

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

= \int \frac{1}{u ^{5}} d u

Apply the Power Rule

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

= - \frac{1}{4 u ^{4}}

Substitute back

u = sin (x)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( 3 x ^{2} + 1 )}{x} d x

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\int sin (c)

\int - e^{- x} cos (e^{- x}) d x

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