What's the integral of $\frac{1}{x^2}\csc^2\left(\frac{1}{x}\right)$?

Question

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.

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 ... — Fred Yang, Apr 29 '16 at 06:39
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? — Karan Singh, Apr 30 '16 at 06:58
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! — Fred Yang, Apr 30 '16 at 13:32

score 4 · Accepted Answer · edited Apr 30 '16 at 16:19

4

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 $\cot(1/x) + C$.

edited Apr 30 '16 at 16:19

zz20s

6,852

answered Apr 30 '16 at 14:21

afailedentertainment

101

What's the integral of $\frac{1}{x^2}\csc^2\left(\frac{1}{x}\right)$?

1 Answers1