# How to integrate $\int 2xe^{x^2-y^2}\cos(2xy)- 2ye^{x^2-y^2}\sin(2xy) \ \mathrm dy$?

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 ?

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

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.

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

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.