4. Integration by Parts
Recall: \(\displaystyle \int u\,dv=u\,v-\int v\,du\) where \(du=\dfrac{du}{dx}\,dx\) and \(dv=\dfrac{dv}{dx}\,dx\)
b1. Choosing the 'Parts'
a. Polynomial Examples
Often, integrals of the form \[ \int x^n\,f(x)\,dx \] can be computed by repeated applications of integration by parts, reducing the power of \(x\) by one at each successive integration. We have already computed three examples. On this page, a few more examples will be considered. The general principle is to select the parts as \[\begin{array}{ll} u=x^n & dv=f(x)\,dx \\ du=nx^{n-1}\,dx \quad & v=F(x) \end{array}\] where \(F(x)\) is an antiderivative of \(f(x)\). This reduction works especially well for integrals of the forms: \[\begin{aligned} &\int x^n e^{ax}\,dx \\ &\int x^n\cos ax\,dx \\ &\int x^n\sin ax\,dx \end{aligned}\] where it is easy to integrate the other factor.
Compute \(\displaystyle \int x^3\sin 2x\,dx.\)
Following the "template", we take: \[\begin{array}{ll} u=x^3 & dv=\sin 2x\,dx \\ du=3x^2\,dx \quad & v=-\,\dfrac{1}{2}\cos 2x \end{array}\] Thus: \[ \int x^3\sin 2x\,dx =-\,\dfrac{1}{2}x^3\cos 2x+\dfrac{3}{2}\int x^2\cos 2x\,dx \] We now apply the parts: \[\begin{array}{ll} u=x^2 & dv=\cos 2x\,dx \\ du=2x\,dx \quad & v=\dfrac{1}{2}\sin 2x \end{array}\] Then \[ \int x^3\sin 2x\,dx =-\,\dfrac{1}{2}x^3\cos 2x +\dfrac{3}{2}\left(\dfrac{1}{2}x^2\sin 2x-\int x\sin 2x\,dx\right) \] Finally, we use the parts: \[\begin{array}{ll} u=x & dv=\sin 2x\,dx \\ du=dx \quad & v=-\,\dfrac{1}{2}\cos 2x \end{array}\] Then \[\begin{aligned} \int x^3\sin 2x\,dx &=-\,\dfrac{1}{2}x^3\cos 2x+\dfrac{3}{4}x^2\sin 2x -\dfrac{3}{2}\left(-\,\dfrac{1}{2}x\cos 2x+\dfrac{1}{2}\int \cos 2x\,dx\right) \\ &=-\,\dfrac{1}{2}x^3\cos 2x+\dfrac{3}{4}x^2\sin 2x+\dfrac{3}{4}x\cos 2x -\dfrac{3}{8}\sin 2x+C \end{aligned}\] Notice that, at each step, we keep the polynomial as \(u\).
We check by differentiating. If \(f=-\,\dfrac{1}{2}x^3\cos 2x+\dfrac{3}{4}x^2\sin 2x +\dfrac{3}{4}x\cos 2x-\dfrac{3}{8}\sin 2x\), then: \[\begin{aligned} f'&=-\,\dfrac{3}{2}x^2\cos 2x+x^3\sin 2x +\dfrac{3}{2}x\sin 2x+\dfrac{3}{2}x^2\cos 2x \\ &\quad+\dfrac{3}{4}\cos 2x-\dfrac{3}{2}x\sin 2x -\dfrac{3}{4}\cos 2x \\ &=x^3\sin 2x \end{aligned}\] which is the function we started with.
Compute the integral \(\displaystyle \int x^3\cos 3x\,dx\).
\(\displaystyle \int x^3\cos 3x\,dx =\dfrac{1}{3}x^3\sin 3x+\dfrac{1}{3}x^2\cos 3x -\dfrac{2}{9}x\sin 3x-\dfrac{2}{27}\cos 3x+C\)
\(\begin{array}{ll} u=x^3 & dv=\cos 3x\,dx \\ du=3x^2\,dx \quad & v=\dfrac{1}{3}\sin 3x \end{array}\) \[ \int x^3\cos 3x\,dx=\dfrac{1}{3}x^3\sin 3x-\int x^2\sin 3x\,dx \] \(\begin{array}{ll} u=x^2 & dv=-\sin 3x\,dx \\ du=2x\,dx \quad & v=\dfrac{1}{3}\cos 3x \end{array}\) \[ \int x^3\cos 3x\,dx =\dfrac{1}{3}x^3\sin 3x+\dfrac{1}{3}x^2\cos 3x-\int \dfrac{2}{3}x\cos 3x\,dx \] \(\begin{array}{ll} u=x & dv=-\,\dfrac{2}{3}\cos 3x\,dx \\ du=\,dx \quad & v=-\,\dfrac{2}{9}\sin 3x \end{array}\) \[\begin{aligned} \int x^3\cos 3x\,dx &=\dfrac{1}{3}x^3\sin 3x+\dfrac{1}{3}x^2\cos 3x-\dfrac{2}{9}x\sin 3x +\int \dfrac{2}{9}\sin 3x\,dx \\ &=\dfrac{1}{3}x^3\sin 3x+\dfrac{1}{3}x^2\cos 3x-\dfrac{2}{9}x\sin 3x -\dfrac{2}{27}\cos 3x+C \end{aligned}\]
We check by differentiating. If \(f =\dfrac{1}{3}x^3\sin 3x+\dfrac{1}{3}x^2\cos 3x -\dfrac{2}{9}x\sin 3x-\dfrac{2}{27}\cos 3x\), then \[\begin{aligned} f'&=x^2\sin 3x+x^3\cos 3x+\dfrac{2}{3}x\cos 3x-x^2\sin 3x \\ &\quad-\dfrac{2}{9}\sin 3x-\dfrac{2}{3}x\cos 3x+\dfrac{2}{9}\sin 3x \\ &=x^3\cos 3x \end{aligned}\]