4. Integration by Parts

Recall:       \(\displaystyle \int u\,dv=u\,v-\int v\,du\)   where   \(du=\dfrac{du}{dx}\,dx\)   and   \(dv=\dfrac{dv}{dx}\,dx\)

a2. Integration by Parts with Multiple Steps

Sometimes you will need to apply integration by parts more than once:

Compute \(\displaystyle \int x^2\cos x\,dx\)

We apply integration by parts with \[\begin{array}{ll} u=x^2 & dv=\cos x\,dx \\ du=2x\,dx \quad & v=\sin x \end{array}\] So: \[ \int x^2\cos x\,dx =x^2\sin x-2\int x\,\sin x\,dx \] Notice that the remaining integral is simpler than the original integral because there is only an \(x\) instead of an \(x^2\). The idea is to keep reducing the power until there are no \(x\)'s left. So to evaluate the second integral, we need to apply integration by parts again, this time with \[\begin{array}{ll} u=x & dv=\sin x\,dx \\ du=\,dx \quad & v=-\cos x \end{array}\] This gives: \[\begin{aligned} \int x^2\cos x\,dx &=x^2\sin x-2\left(-x\cos x+\int \cos x\,dx\right) \\ &=x^2\sin x-2(-x\cos x+\sin x)+C \\ &=x^2\sin x+2x\cos x-2\sin x+C \end{aligned}\]

It is important that in the second step we picked \(u=x\) rather than \(u=\sin x\). Otherwise, we will go in circles as shown here:

Notice that since the polynomial was quadratic, we needed to do integration by parts twice. In the next example, the polynomial is cubic. So we expect three integrations by parts.