Have been a doing a reduction of order ODE problem and this integral comes up at the last step. Not sure how to go about integrating it. The answers give $\cos x^2$ as the answer. Here's the original question: Verify that $u_1=\sin x^2$ is a solution to the equation $$xu''-u'+4x^3u=0$$ and use reduction of order to find a second, linearly independent solution.

I've called the second solution $v$ and as far as I can tell, everything is good with my previous working. The only remaining bit is to integrate $$v'=\frac{Cx}{(\sin x^2)^2}\Leftrightarrow v=\int\frac{Cx}{(\sin x^2)^2}\,\mathrm dx.$$ Integration by parts didn't really help. I think there might be a substitution that I'm missing/forgetting. Thanks.

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    $\begingroup$ Did you try starting with $u=x^2,du=2xdx$? Alternately, you could use $1-\cos^2x^2=sin^2x^2$ and switch to a double angle formula. $\endgroup$ – abiessu Sep 21 '13 at 1:34
  • $\begingroup$ The Maple command $$Student[Calculus1]:-IntTutor(C*x/sin(x^2)^2, x); $$ does the job step by step with explanations. See its output. $\endgroup$ – user64494 Sep 21 '13 at 4:04

In view of the $x$ sitting on top, the substitution $u=x^2$ is natural. So let $u=x^2$. Then $x\,dx=\frac{du}{2}$. We end up with $$\int \frac{C}{2\sin^2 u}\,du.$$ One way to continue is to rewrite as $$\int \frac{C}{2}\csc^2 u\,du.$$ This may not be quite familiar, though integrating its close relative $\sec^2 u$ is familiar. One can verify that our integral is $-\frac{C}{2}\cot u+D$.

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  • $\begingroup$ extra square in the final answer $\endgroup$ – Artem Sep 21 '13 at 2:27
  • $\begingroup$ Thanks! I had hoped this answer was so short as to be trouble-free. $\endgroup$ – André Nicolas Sep 21 '13 at 2:35

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