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How do integrate $x\cdot{\cosh(x^2)}$?

Do i just use integration by parts?

I know that integration by parts is $\int{u\cdot{\mathrm{d}v}} = uv - \int{v\cdot{\mathrm{d}u}}$

Making $v=\frac{1}{2}x^2,\mathrm{d}v=x \ , \ u=\cosh(x^2), \mathrm{d}u=2x \sinh(x^2)$ I get $$ \int{x \cosh(x^2)}\mathrm{d}x=\frac{1}{2}x^2\cosh(x^2)-\int x^3\sinh(x^2)\mathrm{d}x$$ How do i go from here? it seems that another round of integration by parts will only complicate things.

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    $\begingroup$ After 4 months on this site you should know some basic LaTeX. $\endgroup$ – Null Oct 23 '14 at 5:44
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    $\begingroup$ For $x\cosh(x^2)$, use substitution. The function $\cosh(x^2)$ of the title (but not of the question) does not have an elementary antiderivative. $\endgroup$ – André Nicolas Oct 23 '14 at 5:44
  • $\begingroup$ Null you are right i should have looked at the thread about latex ages ago. I actually tried to learn latex before on another site because really hard for that site. In here it is much easier. I did the latex on one of my questions before. $\endgroup$ – Ivan Oct 24 '14 at 5:59
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$$\int x\cosh(x^2)dx$$

substitute $x^2=u$ $$2x dx\Leftarrow\Rightarrow du$$ $$\int \frac{\cosh(u)}{2}du$$ $$\int x\cosh(x^2)dx=\frac{\sinh(u)}{2}+C=\frac{\sinh(x^2)}{2}+C$$ $$\int x\cosh(x^2)dx=\frac{\sinh(x^2)}{2}+C$$

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    $\begingroup$ +some C as integrating constant. $\endgroup$ – Galc127 Oct 23 '14 at 6:15
  • $\begingroup$ @Galc127 Yes, +C as integrating constant $\endgroup$ – user171358 Oct 23 '14 at 6:18
  • $\begingroup$ Thanks a lot for the help =) $\endgroup$ – Ivan Oct 24 '14 at 5:54

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