Integration by parts

The Integration by parts formula

  • Just as the Substitution rule is the method of integration corresponding to the Chain rule, Integration by parts is the method of integration corresponding to the Product rule.
    • Let \(f\) and \(g\) be differentiable functions.
    • Then the product rule says \[ (f(x)g(x))'=f'(x)g(x)+f(x)g'(x). \]
    • This can be rearranged as \[ f(x)g'(x)=(f(x)g(x))'-f'(x)g(x). \]
    • Taking antiderivatives, we get \[ \int f(x)g'(x)\,\mathrm dx=\int (f(x)g(x))'\,\mathrm dx-\int f'(x)g(x)\,\mathrm dx. \]
    • By definition, an antiderivative of the derivative is the original function, therefore we get a first formulation of Integration by parts: \[ \int f(x)g'(x)\,\mathrm dx=f(x)g(x)-\int f'(x)g(x)\,\mathrm dx. \]
  • There is another formulation, which might be easier to remember.
    • Let \(u=f(x)\) and \(v=g(x)\).
    • Then via the Substitution rule, the formula can be changed to \[ \int u\,\mathrm dv=uv-\int v\,\mathrm du. \]

Differentials

  • When working with Integration by parts, it is potentially easier to work with differentials then with derivatives.
    • A differential is an expression like \(\mathrm dv\). You can think of it as the Mathematical object you get when you multiply \(g'(x)=\frac{\mathrm dv}{\mathrm dx}\) with \(\mathrm dx\).
    • In fact, that is all you need to know about differentials:
    • Notation. Let \(g\) be a differentiable function, and let \(v=g(x)\). Then the differential of \(v=g(x)\) is \[ \mathrm dv=g'(x)\mathrm dx. \]
  • Examples.
    • Let \(v=g(x)=x^2+2x-3\).
    • Then we have \(\mathrm dv=g'(x)\mathrm dx=(2x+2)\mathrm dx\).
    • Let \(v=g(x)=\sin3x\).
    • Then we have \(\mathrm dv=3\cos3x\mathrm dx\).
    • Let \(v=g(x)=e^{-2x}\).
    • Then we have \(\mathrm dv=-2e^{-2x}\mathrm dx\).

Integration by parts with a polynomial.

  • 7.1.3. \(\int x\cos5x\,\mathrm dx\).
    • To use the Integration by parts formula, we need to break up the expression \(x\cos 5x\,\mathrm dx\) into a \(u\) part and a \(\mathrm dv\) part.
    • Note that the RHS (right-hand side) of the formula also has an indefinite integral.
    • Therefore, when you separate the expression \(x\cos5x\,\mathrm dx\) into a \(u\) and a \(\mathrm dv\), what you need to think about is that you want to do this in a way that \(\int v\,\mathrm du\) is simpler than \(\int u\,\mathrm dv\).
    • The choice is not always obvious, and you have to be prepared to try out multiple choices before you get a solution.
    • Here, let's take \(u=x\) and \(\mathrm dv=\cos5x\mathrm dx\).
    • Now we have to say what \(\mathrm du\) and \(v\) are.
    • Via the formula, we have \(\mathrm du=f'(x)\mathrm dx=1\cdot\mathrm dx=\mathrm dx\).
    • If \(\mathrm dv=\cos 5x\mathrm dx\), then \(v=g(x)\) is an antiderivative of \(g'(x)=\cos5x\). We can choose \(v=\frac15\sin 5x\).

Integration by parts with a polynomial.

  • Now we are ready to apply the formula: \[ \int x\cos5x\,\mathrm dx=\int u\,\mathrm dv=uv-\int v\,\mathrm du=x\frac15\sin5x-\int\frac15\sin5x\,\mathrm dx. \]
    • Why the choices \(u=x\) and \(\mathrm dv=\cos5x\mathrm dx\) are right is that we know how to calculate the indefinite integral on the right: \[ \int v\,\mathrm du=\int\frac15\sin5x\,\mathrm dx=-\frac{1}{25}\cos5x+C. \]
    • This gives the final answer: \[ \int x\cos5x\,\mathrm dx=\frac15x\sin5x-\int\frac15\sin5x\,\mathrm dx=\frac15x\sin5x+\frac{1}{25}\cos5x+C. \]

Integration by parts with a polynomial.

  • 7.1.6. \(\int(x-1)\sin\pi x\,\mathrm dx\).
    • Once again, we have to separate the expression \(u\,\mathrm dv=(x-1)\sin\pi x\).
    • Let's take \(u=x-1\) and \(\mathrm dv=\sin\pi x\,\mathrm dx\).
    • Then we get \(\mathrm du=\mathrm dx\) and \(v=-\frac{1}{\pi}\cos\pi x\).
    • Therefore, Interation by parts gives \[ \int(x-1)\sin\pi x\,\mathrm dx=-(x-1)\frac{1}{\pi}\cos\pi x+\int\frac{1}{\pi}\cos\pi x\,\mathrm dx=\frac{1-x}{\pi}\cos\pi x+\frac{1}{\pi^2}\sin\pi x+C. \]

Integration by parts with a polynomial.

  • 7.1.7. \(\int(x^2+2x)\cos x\,\mathrm dx\).
    • Let's take \(u=x^2+2x\) and \(\mathrm dv=\cos x\mathrm dx\).
    • Then we get \(\mathrm du=(2x+2)\mathrm dx\) and \(v=\sin x\).
    • Thus, Integration by parts gives \[ \int(x^2+2x)\cos x\,\mathrm dx=(x^2+2x)\sin x-\int(2x+2)\sin x\,\mathrm dx. \]
    • Here, the new thing is that in order to calculate \(\int(2x+2)\sin x\,\mathrm dx\), we need to apply integration by parts again.
    • For that, let \(u=2x+2\) and \(\mathrm dv=\sin x\mathrm dx\).
    • Then we get \(\mathrm du=2\mathrm dx\) and \(v=-\cos x\).
    • That in turn gives \[ \int(2x+2)\sin x\,\mathrm dx=-(2x+2)\cos x+\int2\cos x\mathrm dx=-(2x+2)\cos x+2\sin x+C. \]
    • With this, we can finish the original calculation: \[ \int(x^2+2x)\cos x\,\mathrm dx=(x^2+2x)\sin x-\int(2x+2)\sin x\,\mathrm dx=(x^2+2x)\sin x+(2x+2)\cos x-2\sin x+C. \]
  • From these, you can see that if the integrand is a product of a polynomial and some other function you know how to integrate, then you can make \(u\) the polynomial, and \(\mathrm dv\) given by that other function.

Integration by parts for definite integrals

  • Let \(f\) and \(g\) be functions with domain \([a,b]\), which are differentiable on \((a,b)\).
    • Let's now take definite integrals over \([a,b]\) for the product rule formula: \[ f(x)g'(x)=(f(x)g(x))'-f'(x)g(x). \]
    • Using FTC Part 2, we get the Integration by parts for definite integrals formula: \[ \int_a^bf(x)g'(x)\,\mathrm dx=\left[f(x)g(x)\right]_a^b-\int_a^bf'(x)g(x)\,\mathrm dx. \]
    • It's better to use this version for definite integrals. You can still use \(u\) and \(v\) in your calculations if you prefer.
  • 7.1.23. \(\int_0^{1/2}x\cos\pi x\,\mathrm dx\).
    • We need \(x\cos\pi x=uv'\).
    • We let \(u=x\) and \(v'=\cos\pi x\).
    • Then we have \(u'=1\) and \(v=\frac{1}{\pi}\sin\pi x\).
    • Therefore, we get \[ \int_0^{1/2}x\cos\pi x\,\mathrm dx=\left[x\frac{1}{\pi}\sin\pi x\right]_0^{1/2}-\int_0^{1/2}\frac{1}{\pi}\sin\pi x\,\mathrm dx =\frac{1}{2\pi}+\left[\frac{1}{\pi^2}\cos\pi x\right]_0^{1/2}=\frac{1}{2\pi}-\frac{1}{\pi^2}. \]
  • Exercises. 7.1: 24, 28, 29, 20.

\(\int\ln x\,\mathrm dx\).

  • This is a special case, since it is an elementary function which we can integrate via Integration by parts.
    • Let's write \(\ln x\mathrm dx=u\mathrm dv\).
    • We can put \(u=\ln x\) and \(\mathrm dv=\mathrm dx\).
    • Then we have \(\mathrm du=\frac{1}{x}\mathrm dx\) and \(v=x\).
    • Therefore, we get \[ \int\ln x\,\mathrm dx=x\ln x-\int\frac{x}{x}\mathrm dx=x\ln x-x+C. \]
  • Exercises. 7.1: 10, 11, 15, 26, 30.

Integration by parts with exponential functions, \(\sin\) and \(\cos\)

  • 7.1.18. \(\int e^{-\theta}\cos2\theta\,\mathrm d\theta\).
    • When you want to integrate a product of two functions which are exponential or \(\sin\) or \(\cos\), what's important to remember is to keep performing the same operation on the same term. Let me first show what happens if you do not do this.
    • Let's start out by letting \(u=e^{-\theta}\) and \(\mathrm dv=\cos2\theta\mathrm d\theta\).
    • Then we get \(\mathrm du=-e^{-\theta}\mathrm d\theta\) and \(v=\frac12\sin2\theta\).
    • Therefore, Integration by parts gives \[ \int e^{-\theta}\cos2\theta\,\mathrm d\theta=\frac12e^{-\theta}\sin2\theta+\int\frac12e^{-\theta}\sin2\theta\,\mathrm d\theta. \]
    • WRONG CHOICE We need to perform Integration by parts again. Let's see what happens if we take \(u=\frac12\sin2\theta\) and \(\mathrm dv=e^{-\theta}\).
    • This gives \(\mathrm du=\cos2\theta\mathrm d\theta\) and \(v=-e^{-\theta}\).
    • Thus, the calculation continues as \[ \frac12e^{-\theta}\sin2\theta+\int\frac12e^{-\theta}\sin2\theta\,\mathrm d\theta=\frac12e^{-\theta}\sin2\theta-\frac12e^{-\theta}\sin2\theta+\int e^{-\theta}\cos2\theta\,\mathrm d\theta. \]
    • We've gotten back to the original integral! We've worked for nothing!

Integration by parts with exponential functions, \(\sin\) and \(\cos\)

  • CORRECT CHOICE
    • When performing Integration by parts on \(\int e^{-\theta}\cos2\theta\,\mathrm d\theta\), we had \(u=e^{-\theta}\) and \(\mathrm dv=\cos2\theta\mathrm d\theta\). Therefore, when performing Integration by parts on \(\int\frac12e^{-\theta}\sin2\theta\,\mathrm d\theta\), we need to put \(u=e^{-\theta}\) and \(\mathrm dv=\frac12\sin2\theta\).
    • (Actually, it's OK to pull the \(\frac12\) out of the integral, I just keep it there to make the argument clearer.)
    • We get \(\mathrm du=-e^{-\theta}\,\mathrm dx\) and \(v=-\frac14\cos2\theta\).
    • Therefore, the calculation continues as \[ \frac12e^{-\theta}\sin2\theta+\int\frac12e^{-\theta}\sin2\theta\,\mathrm d\theta=\frac12e^{-\theta}\sin2\theta-\frac14e^{-\theta}\cos2\theta-\int\frac14e^{-\theta}\cos2\theta\,\mathrm d\theta. \]
    • We have gotten \[ \int e^{-\theta}\cos2\theta\,\mathrm d\theta=\frac12e^{-\theta}\sin2\theta-\frac14e^{-\theta}\cos2\theta-\frac14\int e^{-\theta}\cos2\theta\,\mathrm d\theta. \]
    • Note that we have the same integral on the two sides of the equation.
    • Therefore, we can rearrange it as \[ \frac54\int e^{-\theta}\cos2\theta\,\mathrm d\theta=\frac12e^{-\theta}\sin2\theta-\frac14e^{-\theta}\cos2\theta. \]
    • That is, the final answer is \[ \int e^{-\theta}\cos2\theta\,\mathrm d\theta=\frac25e^{-\theta}\sin2\theta-\frac15e^{-\theta}\cos2\theta. \]

Exercises.

  • 7.1: 17, 9, 12, 37, 40, 41, 48, 52.