# Ceci n'est pas un pipe

## 23 October 2013

Today in Calculus II:

$\int x\sin x\cos x\,dx$

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:

$\sin 2\theta=2\sin\theta\cos\theta$

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

$\frac{1}{2}\sin2\theta=\sin\theta\cos\theta$

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

\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.

\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.