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.

  • 5
    $\begingroup$ After 4 months on this site you should know some basic LaTeX. $\endgroup$ – Null Oct 23 '14 at 5:44
  • 2
    $\begingroup$ 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

$$\int x\cosh(x^2)dx$$

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

  • 1
    $\begingroup$ +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

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.