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It's known that $\int\csc^2(x)dx = -\cot(x) + C$, but I don't know how to integrate $\int\frac{1}{x^2}\csc^2(\frac{1}{x})dx$. Can you help? Answer to integral $\int_{2}^b\frac{1}{x^2}\csc^2(\frac{1}{x})dx$ also works.

Answer of this question may give some hints to solution of another question I posted here.

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  • $\begingroup$ See here? $\endgroup$
    – S.C.B.
    Commented Apr 29, 2016 at 6:26
  • $\begingroup$ Thanks for comments! I don't know if there's a closed form, and the walframalpha.com link cannot produce result for this integral, so I'm seeking help ... $\endgroup$
    – Fred Yang
    Commented Apr 29, 2016 at 6:39
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    $\begingroup$ @FredYang The new integral is a simple $u$ sub. $\endgroup$
    – MathematicsStudent1122
    Commented Apr 30, 2016 at 6:54
  • $\begingroup$ I don't know why wolfram alpha says it has no solution.I can put $\frac {1}{x}$ =$t$ to get the integral.Is there anything wrong in my logic? $\endgroup$
    – Karan Singh
    Commented Apr 30, 2016 at 6:58
  • $\begingroup$ The original question about integral $\csc^2\left(\frac{1}{x}\right)$ had no solution on wolfram, but updated question about integral of $\frac{1}{x^2}\csc^2\left(\frac{1}{x}\right)$ has a solution now. Thanks for your comments! $\endgroup$
    – Fred Yang
    Commented Apr 30, 2016 at 13:32

1 Answer 1

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Notice that the term $1/x^2$ is the negative derivative of the function's argument $1/x$. Set $u=1/x$, $du = -1/x^2$. $-du=1/x^2 dx$. The new integral is of $-\csc^2(u)du$. Integrate that to get $cot(u) + C$, and substitute $u$ back into the expression to get $\cot(1/x) + C$.

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