Cogito
AP Calculus BC · Chapter 6 · Lesson 3
Choosing an Integration Technique
Recognising which tool the integral wants.
12 problems · about 21 minutes · FUN-6.D
What this lesson teaches
The student selects an appropriate integration technique for a given integral.
- Try the cheapest technique first: basic form, then substitution, then parts or partial fractions.
- Substitution needs a function and its derivative both present.
- Every answer is checkable by differentiating.
Warm Up
Straightforward practice. Get the method working first.
5 problemsx times sin x. Which technique? 1 substitution, 2 parts.
Answer 2
Why Integration by parts.
What signals substitution?
Answer A function and its derivative are both present.
Why The inner derivative appearing as a factor.
x times eˣ. Which technique? 1 substitution, 2 parts.
Answer 2
Why The derivative of x is not eˣ.
x times e^(x²). Which technique? 1 substitution, 2 parts.
Answer 1
Why The inner derivative is present.
1 over ((x−1)(x−2)). Which technique? 1 partial fractions, 2 parts.
Answer 1
Why A factorable denominator.
Build It Up
The same ideas with more to keep track of.
3 problemsln x on its own. Which technique? 1 substitution, 2 parts.
Answer 2
Why Take u = ln x and dv = dx.
Can every integration answer be checked by differentiating? 1 yes, 0 no.
Answer 1
Why The operations are inverse.
Which is cheapest to try first? 1 substitution, 2 partial fractions.
Answer 1
Why Less setup.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsSort each integral by the technique it wants.
Answer Substitution: x times e to the x squared, 2x times (x squared plus 1) cubed · Integration by parts: x times sin x, x times e to the x
Why Ask whether the inner derivative is present as a factor.
2x times (x² + 1)³. Which technique? 1 substitution, 2 parts.
Answer 1
Why The inner derivative is right there.
The Signal: A function and its derivative are both present. Which technique? 1 substitution, 2 parts.
Answer 1
Why Substitution.
The Rational: A rational function with a factorable denominator. Which technique? 1 partial fractions, 2 parts.
Answer 1
Why Partial fractions.