What is the integral of 1/(x^3)csc^4(1/(x^2)) ?

The integral of 1/(x^3)csc^4(1/(x^2)) is 1/6 csc^3(1/(x^2))cos(1/(x^2))+1/3 cot(1/(x^2))+C

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integral of 1/(x^3)csc^4(1/(x^2))

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

\frac{1}{6} csc^{3} (\frac{1}{x ^{2}}) cos (\frac{1}{x ^{2}}) + \frac{1}{3} cot (\frac{1}{x ^{2}}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{csc ^{4} ( u )}{2} d u

= \int - \frac{csc ^{4} ( u )}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int csc^{4} (u) d u

Apply Integral Reduction

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

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

= - \frac{1}{2} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} (- cot (u)))

Substitute back

u = \frac{1}{x ^{2}}

= - \frac{1}{2} (- \frac{cos ( \frac{1}{x ^{2}} ) csc ^{3} ( \frac{1}{x ^{2}} )}{3} + \frac{2}{3} (- cot (\frac{1}{x ^{2}})))

Simplify

- \frac{1}{2} (- \frac{cos ( \frac{1}{x ^{2}} ) csc ^{3} ( \frac{1}{x ^{2}} )}{3} + \frac{2}{3} (- cot (\frac{1}{x ^{2}}))) : \frac{1}{6} csc^{3} (\frac{1}{x ^{2}}) cos (\frac{1}{x ^{2}}) + \frac{1}{3} cot (\frac{1}{x ^{2}})

= \frac{1}{6} csc^{3} (\frac{1}{x ^{2}}) cos (\frac{1}{x ^{2}}) + \frac{1}{3} cot (\frac{1}{x ^{2}})

Add a constant to the solution

= \frac{1}{6} csc^{3} (\frac{1}{x ^{2}}) cos (\frac{1}{x ^{2}}) + \frac{1}{3} cot (\frac{1}{x ^{2}}) + C

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What is the integral of 1/(x^3)csc^4(1/(x^2)) ?
The integral of 1/(x^3)csc^4(1/(x^2)) is 1/6 csc^3(1/(x^2))cos(1/(x^2))+1/3 cot(1/(x^2))+C