3
$\begingroup$

Find the value of: $$\int 2xe^{x^2-y^2}\cos(2xy)- 2ye^{x^2-y^2}\sin(2xy) \ \mathrm dy$$

i have tried using integration by parts but it doesn't seem to work. How would I go about this integral ?

$\endgroup$
  • $\begingroup$ Is this the original exercise? $\endgroup$ – mickep May 13 '17 at 17:03
  • 2
    $\begingroup$ Knowing the domain of integration might prove helpful. $\endgroup$ – Jonathan Y. May 13 '17 at 17:05
  • 1
    $\begingroup$ It is integration with respect to y $\endgroup$ – april analysis May 13 '17 at 17:13
  • 1
    $\begingroup$ Well, that settles that then. $\endgroup$ – tilper May 13 '17 at 17:13
  • 1
    $\begingroup$ @tilper yes it is an indefinite integral $\endgroup$ – april analysis May 13 '17 at 17:16
3
$\begingroup$

Since

$$ \frac{d}{dy}\left[e^{x^2-y^2}\sin(2xy)\right]=- 2ye^{x^2-y^2}\sin(2xy)+2xe^{x^2-y^2}\cos(2xy)$$

then

$$ \int 2xe^{x^2-y^2}\cos(2xy)- 2ye^{x^2-y^2}\sin(2xy)\,dy= e^{x^2-y^2}\sin(2xy)+c$$

Note: Because of the fact that the integrand appeared to be the result of application of the product rule for derivatives, it seemed appropriate to investigate the possibility.

$\endgroup$
3
$\begingroup$

$$\int \left(2xe^{x^2-y^2}\cos 2xy - 2ye^{x^2-y^2}\sin 2xy\right) \, dy = \int 2xe^{x^2-y^2} \cos 2xy \, dy - \int 2ye^{x^2-y^2} \sin 2xy \, dy $$

Let's integrate that first one on the RHS by parts. We'll take $u = e^{x^2-y^2}$. So then $du = -2ye^{x^2-y^2} \, dy$, $dv = 2x \cos 2xy \, dy$, and $v = \sin 2xy$. (It may not be clear at first, but note that $v$ can easily be obtained from $dv$ with a simple substitution integral. Substitute $t = 2xy$ into $\int 2x \cos 2xy \, dy$ to get $v = \sin 2xy$.)

With integration by parts, we have \begin{align*} \int 2xe^{x^2-y^2} \cos2xy \, dy &= e^{x^2-y^2}\sin 2xy - \int \sin (2xy) \cdot \left(-2y e^{x^2-y^2}\right) \, dy\\[0.3cm] &= e^{x^2-y^2}\sin 2xy + \int 2y e^{x^2-y^2} \sin 2xy \, dy\\[0.3cm] \end{align*}

Now use this with the very first line of my post to get a great cancellation, giving you the desired result immediately.

$\endgroup$

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.