What is the integral of x^{-4}(csc^2(x^{-3}+1)) ?

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

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integral of x^{-4}(csc^2(x^{-3}+1))

\int x^{- 4} (csc^{2} (x^{- 3} + 1)) d x

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

Solution steps

\int x^{- 4} csc^{2} (x^{- 3} + 1) d x

Apply u-substitution

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

= \int - \frac{csc ^{2} ( u )}{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 csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= - \frac{1}{3} (- cot (u))

Substitute back

u = x^{- 3} + 1

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

Simplify

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

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

Add a constant to the solution

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

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