Calc II Lesson 4

Professor Abdul-Quader

L4 Integration by Parts / Trig Integrals

Questions?

So far:

  • FTC / Integration
  • Substitution
  • Inverse Trig Functions

Reminders

  • MyOpenMath Homework due tomorrow.
  • Written homework due Monday in class.
  • Quiz on Monday!

Integration by Parts

  • Substitution: “Reverse chain rule”
  • Integration by parts: “Reverse product rule”

\[(fg)^\prime = f^\prime g + fg^\prime\]

Reverse? \(\int f^\prime(x)g(x) dx + \int f(x) g^\prime(x) dx = f(x) g(x) + C\)

Reverse product rule

\[\int f^\prime(x)g(x) dx + \int f(x) g^\prime(x) dx = f(x) g(x) + C\]

Or:

\[\int f(x) g^\prime(x) dx = f(x) g(x) - \int f^\prime(x) g(x) dx + C\]

Substitute:

  • \(u = f(x)\), \(du = f^\prime(x) dx\)
  • \(v = g(x)\), \(dv = g^\prime(x) dx\)

Then:

\[\int u dv = uv - \int v du\]

Integration by parts formula!

Example

\(\int x e^x dx\). Must pick \(u\) and \(dv\).

  • Let \(u = x\), \(dv = e^x dx\).
  • Then \(du = dx\), \(v = e^x\)
  • \(uv - \int v du = xe^x - \int e^x dx\)
  • Or just \(xe^x - e^x + C\)

Q: What if we chose \(u = e^x\), \(dv = x dx\)?

Hmm…

  • \(u = e^x\), \(dv = x dx\)
  • \(du = e^x dx\), \(v = \frac{x^2}{2}\)
  • \(uv - \int vdu = \frac{x^2}{2} e^x - \int \frac{x^2}{2} e^x dx\)
  • More complicated than original problem!
  • Pick \(u\) and \(v\) carefully!

Guidance

  • Picking \(u\) and \(v\) takes some intuition.
  • Generally:
    • \(u\) should be easy to differentiate.
    • \(du\) should be simpler than \(u\)
    • \(dv\) should be easy to integrate
    • \(v\) should not be much more complicated than \(u\)

More guidance

  • For \(u\), prefer:
    • logarithms (\((\ln|x|)^\prime = \frac{1}{x}\))
    • inverse trig functions
    • polynomials / algebraic functions
  • For \(dv\):
    • trig (sin / cos)
    • exponential functions

For \(u\): LIPET. Logs > Inverse Trig > Polynomials > Exponentials > Trig.

Exercises

  1. \(\int x \ln(x) dx\)
  2. \(\int \frac{\ln(x)}{x^2} dx\)

Solution (1)

\[\int x \ln(x) dx\]

  • Let \(u = \ln(x)\), \(dv = x dx\)
  • \(du = \frac{1}{x} dx\), \(v = \frac{x^2}{2}\)
  • \(uv - \int v du = \frac{x^2}{2} \ln(x) - \int \frac{x}{2} dx\)
  • Or \(\frac{x^2}{2}\ln(x) - \frac{x^2}{4} + C\)

Solution (2)

\[\int \frac{\ln(x)}{x^2} dx\]

  • \(u = \ln(x)\), \(dv = \frac{dx}{x^2}\)
  • \(du = \frac{1}{x} dx\), \(v = -\frac{1}{x}\)
  • \(-\frac{\ln(x)}{x} - \int -\frac{1}{x^2} dx\), or
  • \(-\frac{\ln(x)}{x} + \int \frac{1}{x^2} dx\)
  • Final answer: \(-\frac{\ln(x)}{x} - \frac{1}{x} + C\)

Rest

On your own, take a look at the rest of today’s notes / videos. This covers:

  • Harder integration by parts examples.
  • Integrals of powers of trig functions (like \(\sin^2(x)\) or \(\cos^4(x)\)).
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