Ceci n'est pas un pipe

23 October 2013

Today in Calculus II:

∫xsin⁡xcos⁡x dx\int x\sin x\cos x\,dx

You’ll notice that there’s three symbols in this term, which makes integration by parts impossible. But there’s not actually three symbols there. No, like Magritte before me, I must tell you that this is not a pipe – or in this case a three-symbol term. It’s actually a very compact little two-symbol term.

Let me remind you of the Double-Angle Formula:

sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta=2\sin\theta\cos\theta

Therefore, we could easily rewrite that to be just as true:

12sin⁡2θ=sin⁡θcos⁡θ\frac{1}{2}\sin2\theta=\sin\theta\cos\theta

This is easily accomplished using integration by parts, which seems to be my theme for this month.

∫12xsin⁡2x dx=12[12xcos⁡2x−∫−12cos⁡2x dx]u=xdv=sin⁡2x dxdu=dxv=−12cos⁡2x\begin{array}{cl} \int\frac{1}{2}x\sin2x\,dx&=\frac{1}{2}\left[\frac{1}{2}x\cos2x-\int\frac{-1}{2}\cos2x\,dx\right]\\ \begin{aligned} u&=x&dv&=\sin2x\,dx\\ du&=dx&v&=\frac{-1}{2}\cos2x \end{aligned} \end{array}

The remainder of which is easily integrated and simplified.

12[12xcos⁡2x−∫−12cos⁡2x dx]=12[−12xcos⁡2x+14sin⁡2x]=18sin⁡2x−14xcos⁡2x+C\begin{aligned} \frac{1}{2}\left[\frac{1}{2}x\cos2x-\int\frac{-1}{2}\cos2x\,dx\right]&=\frac{1}{2}\left[\frac{-1}{2}x\cos2x+\frac{1}{4}\sin2x\right]\\ &=\frac{1}{8}\sin2x-\frac{1}{4}x\cos2x+C\end{aligned}

This is a solution for §7.1 Exercise #21 from Calculus: Early Transcendentals.