1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $-\frac{C}{2}\cot u+D$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.