In how many different ways can you find this? \[\displaystyle{\int {\cos (x) \sin (x)} \, dx}\]

The integrand is familiar from the identity \(\sin(2x) \equiv 2\sin(x) \cos(x)\) so we could use this to help us.

\[\displaystyle{\int {\cos (x) \sin (x)} \, dx} = \displaystyle{\int \dfrac{1}{2} {\sin (2x)} \, dx} = -\dfrac{1}{4} \cos(2x) + C\]

If we use the substitution \(u=\cos(x)\), then \(\dfrac{du}{dx} = -\sin(x)\).

Therefore, \[\displaystyle{\int {\cos (x) \sin (x)} \, dx} = \displaystyle{\int {-u} \, du} = -\dfrac{1}{2} \cos^2 (x) + C\]

Is this the only possible substitution we could use?

We have a product of two different functions, so integration by parts may be useful. \[\int uv' \, dx = uv - \int u'v \, dx\]

We let \(u=\sin(x)\) and \(v'=\cos(x)\), so that \(u'=\cos x\) and \(v=\sin x\).

\[\displaystyle{\int {\cos (x) \sin (x)} \, dx}= \sin^2(x) - \displaystyle{\int {\cos (x) \sin (x)} \, dx}\]

The integral on the right hand side is the same as our original. Has this helped?

Let’s define \[\displaystyle{\int {\cos (x) \sin (x)} \ dx} = I.\]

Then we have \[I = \sin^2(x) -I +c.\]

Making

\(I\) the subject, we have

\[\begin{align*}
2I &= sin^2(x) +c\\
\implies\quad I&= \dfrac{1}{2} \sin^2(x) + C.
\end{align*}\]
Would the answer have been different if we had swapped the \(u\) and \(v'\)?

We’ve integrated the function in three different ways and seem to have three different answers. What has happened?