0
$\begingroup$

I have a really vague integration problem. It's some substitution and then integration by parts maybe. I got this from a friend, but it seems it's unlikely​ to be solved.

Problem Now my question is, how to approach such big problems?

I reduced it to $$\displaystyle{\int \bigg(3x(2x-x\cos x - \sin x)\sqrt{1-\frac{\sin x}{x}}} \bigg)\mathrm{d}x$$

How to proceed from here?

$\endgroup$
  • $\begingroup$ If you take $u=x-\sin x$, then $du=2\sin^2\frac{x}{2}dx$ looks promising, but I can't get rid of those extra $x$ terms. $\endgroup$ – orion Jun 29 '17 at 19:35
  • $\begingroup$ @orion Neither can I, unfortunately. $\endgroup$ – Mathejunior Jun 29 '17 at 19:38
  • $\begingroup$ it was the right direction, just needed one more step. See the solution below. $\endgroup$ – orion Jun 29 '17 at 19:44
4
$\begingroup$

I think I got it! Consider your simplified version and try to regroup terms to make it look more organized:

$$\int 3x(2x-x\cos x-\sin x )\sqrt{1-\frac{\sin x}{x}}dx$$ $$=\int 3\sqrt{x}(x(1-\cos x) +(x-\sin x))\sqrt{x-\sin x}dx$$ Now start recognition of similar terms. Let's define $$u=x-\sin x$$ Observe $du=(1-\cos x)dx$. The integral becomes: $$=\int 3(x(du/dx) +u)\sqrt{ux}dx=\int 3(xdu+udx)\sqrt{ux}$$ But now you see how the parenthesis looks like a total derivative of $ux$. This leads us to the conclusion, that $u$ is not a good variable substitution, but $ux$ is! Define $t=ux=x^2-x\sin x$ and you get $$\int 3 \sqrt{t}dt$$ You can take it from here.

$\endgroup$
  • $\begingroup$ Very slick! +1 $$ $\endgroup$ – Harry Jun 29 '17 at 19:50
  • $\begingroup$ I accept your answer. Thanks a lot. And the final answer is $2((x^2-x \sin x)^\frac{3}{2})$. Is that correct? $\endgroup$ – Mathejunior Jun 30 '17 at 5:32
  • $\begingroup$ @Mathbg that's correct. $\endgroup$ – orion Jun 30 '17 at 13:24
-1
$\begingroup$

enter image description here

Easier way to solve this problem

$\endgroup$

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.