Calculus 1-The Substitution Rule, guaranteed 100% Pass

1


The Substitution Rule


Not every differentiable function has an elementary antiderivative. For example,
there is no elementary function for ∫ sin(𝑥 2 ) 𝑑𝑥. We know a handful of basic
antiderivatives (e.g. ∫ 𝑥 𝑛 𝑑𝑥, 𝑛 ≠ −1, ∫(𝑠𝑖𝑛𝑎𝑥)𝑑𝑥, etc.). So far we have tried
to turn all of our indefinite integral problems into sums, differences, and constant
multiples of these basic antiderivatives. In this section we will learn to make
substitutions in indefinite integrals that will turn some integrands into sums,
differences, and constant multiples of the basic antiderivatives we know.


Suppose we have a composite function 𝐹(𝑔(𝑥)), where 𝐹(𝑥) is an antiderivative
of 𝑓(𝑥), i.e. 𝐹 ′ (𝑥) = 𝑓(𝑥) and ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶. [ For example, 𝐹(𝑥) = 𝑥 4 ,
𝑔(𝑥) = 𝑥 2 + 1, and 𝑓(𝑥) = 4𝑥 3 . So 𝐹(𝑔(𝑥)) = (𝑥 2 + 1)4 .] By the chain rule:


𝑑
𝐹(𝑔(𝑥 )) = 𝐹 ′ (𝑔(𝑥 ))𝑔′ (𝑥 ) = 𝑓(𝑔(𝑥 ))𝑔′(𝑥).
𝑑𝑥

Thus we have ∫ 𝑓(𝑔(𝑥 ))𝑔′ (𝑥 )𝑑𝑥 = 𝐹(𝑔(𝑥 )) + 𝐶.



In other words, faced with an integral that looks like ∫ 𝑓(𝑔(𝑥 ))𝑔′ (𝑥 )𝑑𝑥 (for
example ∫ 4(𝑥 2 + 1)3 (2𝑥 )𝑑𝑥) we can make a substitution

𝑢 = 𝑔(𝑥)
𝑑𝑢 = 𝑔′ (𝑥 )𝑑𝑥
and ∫ 𝑓(𝑔(𝑥 ))𝑔′ (𝑥 )𝑑𝑥 becomes ∫ 𝑓 (𝑢)𝑑𝑢 = 𝐹 (𝑢) + 𝐶 = 𝐹(𝑔(𝑥 )) + 𝐶.

, 2


Ex. Evaluate ∫ 4(𝑥 2 + 1)3 (2𝑥 )𝑑𝑥.


Let 𝑢 = 𝑥2 + 1
𝑑𝑢 = 2𝑥𝑑𝑥
So the integral becomes:

∫ 4(𝑥 2 + 1)3 (2𝑥 )𝑑𝑥 = ∫ 4𝑢3 𝑑𝑢
= 𝑢4 + 𝐶
= (𝑥 2 + 1)4 + 𝐶
𝑑
Notice ((𝑥 2 + 1)4 + 𝐶 ) = 4(𝑥 2 + 1)3 (2𝑥 ).
𝑑𝑥




Ex. Evaluate ∫ 3𝑥 2 (𝑥 3 + 4)6 𝑑𝑥.


Let 𝑢 = 𝑥 3 + 4

𝑑𝑢 = 3𝑥 2 𝑑𝑥
Now substituting into the integral:
1
∫ 𝑢6 𝑑𝑢 = 7 𝑢7 + 𝐶
1
= (𝑥 3 + 4)7 + 𝐶.
7